trigonometry
limit
calculus
calculus computations

Suppose $ x_{n\plus{}1} \equal{} \frac{x_{n} \plus{} y_{n}}{2}$ and $ y_{n\plus{}1} \equal{} \sqrt{y_{n} \times x_{n\plus{}1}}$, where $ x_{1} \equal{} \cos \theta, y_{1} \equal{} 1$. Calculate $ \lim_{n \to \infty} x_{n}$.

We can assume $ 0\leq \theta\leq\pi$ to attain the full range for $ \cos\theta$. Clearly, $ x_2\equal{}\cos^2\frac{\theta}{2},y_2\equal{}\cos\frac{\theta}{2}$, in particular, $ \forall n>1: y_n\geq x_n\geq 0$. The special case $ \theta \equal{}0$ yields $ \forall n: x_n\equal{}y_n\equal{}1$ and $ \theta\equal{}\pi$ yields $ \forall n>1: x_n\equal{}y_n\equal{}0$. So, from now on let $ 0<\theta<\pi$ and note, that $ \forall n>1: x_{n\plus{}1}\equal{}0$ or $ y_{n\plus{}1}\equal{}0$ implies $ x_n\equal{}y_n\equal{}0$, i.e. we can assume $ \forall n>1: x_ny_n\not\equal{}0$. Since $ y_{n\plus{}1}^2\minus{}x_{n\plus{}1}^2\equal{}\frac{y_n^2\minus{}x_n^2}{4}$ we can focus on the series $ y_n$ instead of $ x_n$ with regard to limit analysis. Denote $ R_n\equal{}\frac{y_{n\plus{}1}}{y_n}$ and observe, that $ 2R_{n\plus{}1}^2\equal{}2\frac{x_{n\plus{}2}y_{n\plus{}1}}{y_{n\plus{}1}^2}\equal{}1\plus{}\frac{x_{n\plus{}1}y_{n\plus{}1}}{y_{n\plus{}1}^2}\equal{}1\plus{}R_n$. Since $ R_1\equal{}\cos\frac{\theta}{2}$, we conclude more generally $ R_n\equal{}\cos\frac{\theta}{2^n}$. This yields $ y_{N\plus{}1}\equal{}\frac{y_{N\plus{}1}}{y_1}\equal{}\prod_{n\equal{}1}^NR_n\equal{}\prod_{n\equal{}1}^N\cos\frac{\theta}{2^n}\equal{}2^{\minus{}N}\prod_{n\equal{}1}^N\frac{\sin\frac{\theta}{2^{n\minus{}1}}}{\sin\frac{\theta}{2^{n}}}\equal{}\frac{\sin\theta}{2^N\sin\frac{\theta}{2^N}}$.

This leads to $ \lim y_N\equal{}\lim x_N\equal{}\frac{\sin\theta}{\theta}$ (this includes the special cases according to Le'Hospital).

Yes, the answer is $ \frac {\sin \theta}{\theta}$.

Since we have $ y_{n \plus{} 1} \equal{} y_n\cos \frac {\theta}{2^{n \plus{} 1}}$. $ \therefore y_{n \plus{} 1}\sin \frac {\theta}{2^{n \plus{} 1}} \equal{} y_n\cdot \frac {1}{2}\left(2\cdot \frac {\theta}{2^{n \plus{} 1}}\right)$

$ \therefore 2^{n \plus{} 1}y_{n \plus{} 1}\sin \frac {\theta}{2^{n \plus{} 1}} \equal{} 2^ny_n\sin \frac {\theta}{2^n}$, yielding $ 2^{n \plus{} 1}y_{n \plus{} 1}\sin \frac {\theta}{2^{n \plus{} 1}} \equal{} 2y_1\sin \frac {\theta}{2} \equal{} \sin \theta$.

$ \therefore y_{n \plus{} 1} \equal{} \sin \theta /\left( \frac {\theta \sin \frac {\theta}{2^{n \plus{} 1}}}{\frac {\theta}{2^{n \plus{} 1}}}\right)$

The desired limit can be obtained as $ \frac {\sin \theta}{\theta}$ without using l'Hopital's theorem.

Note that it is possible to [url=http://en.wikipedia.org/wiki/Arithmetic-geometric_mean#Properties]generalize[/url].

I suppose $ \theta \equal{}0$ requires Le'Hospital


modular arithmetic
floor function
AMC
AIME
articles
function
number theory

Agh, Euler's theorem fails for relatively composite numbers!

[hide]We'll separate $x\equiv2006^{2007}\pmod{2008}$ into $2006^{2007}\pmod{8}$ and $2006^{2007}\pmod{251}$. The first gives $x\equiv0\pmod{8}$ so we'll worry about the second. But the second gives $x\equiv-2^{7}\equiv-128\equiv 123\pmod{251}$. Thus we have $8k\equiv 123\mod 251$. Therefore, we have $k\equiv235\pmod{251}\Longrightarrow k\equiv235\cdot8=1880\pmod{2008}$. And our answer would then be $\boxed{188}$.[/hide]

I got [hide="that same answer"] 1880 [/hide] without splitting it up at all.  When do you [b]need[/b] to split it up and when does it work anyway?

[b]Shorter Solution:[/b]

$GCD(2,2008)=2$, so the period of repetition of $2^{k}$ after $k=1$ is $\varphi (2008/2)= \varphi (1004)= \varphi ((4)(251))= \varphi (4) \varphi (251)= (2)(250) = 500$

$2006^{2007}\equiv (-2)^{2007}\equiv-2^{2007}\equiv-2^{7}\equiv-128 \equiv 1880\pmod{2008}$, so $\boxed{188}$

[b]Remark:[/b]

Saying that the result is divisible by 10 will imply that you have $\equiv-2^{k}$ where $2^{k}$ ends in $8$. The only possibilities for this are $2^{3}= 8$ and $2^{7}= 128$, so it's become a two-choice problem with almost no work.

It should at least say $\lfloor \text{stuff}/10 \rfloor$ instead of $\text{stuff}/10$

If you end up actually making a mock-AIME, look out for these things, because they really kill the whole point of the problem.

[quote="miyomiyo"]Compute the numerical value of
$\frac{2006^{2007}(\text{mod}2008)}{10}$[/quote]

FWIW - You have to be  a little careful, question as worded above is  not clear, for example any multiple of 2008 can be added to the 1880 and all these value will be congruent to  1880 . (when this is divided by 10, you will not have an unique value - you may not even get an integer)   . IOW, you may like to  use   words  like   " remainder  when divided by 2008"

The Zuton Force could you please explain a few things to me about your answer? If you dont want to explain thats fine but could you atleast point me in a direction to understand?

The whole phi thing, how does that work? I realize that you get 500, and I was wondering what that meant until I noticed that $-2^{2007}$ becomes $-2^{7}$ because $7 \equiv 2007 \pmod{500}$.
Atleast I think thats why..

Thanks for the help  :oops:

You reasoning is correct ...

For $\phi(n)$ you can look up  , say wiki  article   [url]http://en.wikipedia.org/wiki/Euler's_totient_function[/url]
Just check out the section, "Computing Euler's function"

Basically the theorem says $a^{\phi(n)}\equiv 1 \bmod n$ as long as $(a,n)$ do not have any common factors.

(Of course, for prime $n$ $\phi(n)$ is just (n-1).   if n is not prime and is product  of, say p and q (p and q have no common factors) then $\phi(n) = \phi(p) \phi(q)$

So $2^{250}\equiv 1 \bmod 251$

Of course, you don't have to go as far as 500 (or even 250) ..In fact you  can figure out that  $2^{50}\equiv 1 \bmod 251$  directly if you happen to notice that $2^{8}= 256 = 251+5$  and $2*5^{3}= 250 = 251-1$

[hide]All mod 251
$-1 \equiv 250 \equiv 2 \cdot 5^{3}$
But $5 \equiv 2^{8}$   (remember $2^{8}= 256 \equiv 5$
 so
$-1 \equiv 2 \cdot (2^{8})^{3}\equiv 2^{25}\implies 2^{50}\equiv 1$ :) [/hide]

[

Perhaps it'd be better if instead of divided by ten, simply: what is the answer modulo 1000, like most AIME problems.

But we're already taking the remainder when divided by 2008!

Yes. Uh, take the remainder modulo 2008 and just ask for the last three digits... right? I think that's what I said.


calculus
integration
LaTeX
trigonometry
calculus computations

[quote="Alpha_new"]Hello, could you please help me to compute such definite integral

Int from 0 to pi/2 1/( 1 + a/sin^2(x) ) dx

BTW, how to do so nice picteres of integrals? I am the first time here.[/quote]You can use Latex to make your text look nicer.  There's a help file on this site, but basically you just put $\$$ symbols around your code; for example, when I put $\$$ symbols around the following code:
x^2+2x+1=(x+1)^2
it looks like this:
$x^2+2x+1=(x+1)^2$.
The code in this case would be something like this:
\int_{0}^{\pi/2}\frac{1}{1+\frac{a}{\sin(x)^2}}dx
which looks like this:
$\int_{0}^{\pi/2}\frac{1}{1+\frac{a}{\sin(x)^2}}dx$

Thank you for advise!!! I will use it since now :)
But anyway, do you know how to solve it? Any suggestions?
I have tried already substitution with sinus and tangens, but it doesn't help. What else?

Waiting for your response!!!

$\frac{1}{1+\frac{a}{\sin^2{x}}}=\frac{\sin^2{x}}{\sin^2{x}+a}=1-\frac{a}{\sin^2{x}+a}$, so the integral $\int_{0}^{\pi/2}\frac{1}{1+\frac{a}{\sin^2{x}}}dx=\int_{0}^{\pi/2}dx-a\int_{0}^{\pi/2}\frac{1}{\sin^2{x}+a}dx$, so all you really need to do is find $\int_{0}^{\pi/2}\frac{1}{\sin^2{x}+a}dx$.  I'll hide part of the answer below:
[hide]$\int\frac{1}{\sin^2{x}+a}dx=\frac{\arctan(\tan{x}\sqrt{\frac{a+1}{a}})}{\sqrt{a(a+1)}}$, except when $a=0$ or $a=-1$.[/hide]By the way, you can see the actual code by hovering your mouse over the image.

Thank you very much!
It was not too difficult :)


[quote="joejia"]Find the integer closest to $\sqrt[3]{700}$.[/quote]

[hide]From memory

$8^3=512$

$9^3=729$

So 9 must be the closest[/hide]

[quote="joejia"]Find the integer closest to $\sqrt[3]{700}$.[/quote]

$8^3=512$

$10^3=1000$

$\text{So } 9^3 \approx 700$


Let $ a,b,c\geq0$ s.t. $ a^3\plus{}b^3\plus{}c^3\equal{}3$.Prove that:
$ a^4b^4\plus{}b^4c^4\plus{}c^4a^4\leq3$

[quote="Inequalities Master"]Let $ a,b,c\geq0$ s.t. $ a^3 \plus{} b^3 \plus{} c^3 \equal{} 3$.Prove that:
$ a^4b^4 \plus{} b^4c^4 \plus{} c^4a^4\leq3$[/quote]
$ a^4b^4 \plus{} b^4c^4 \plus{} c^4a^4\leq\frac{\sum{a^3b^3(a^3\plus{}b^3\plus{}1)}}{3}$
Let $ x\equal{}a^3,y\equal{}b^3,z\equal{}c^3$ so the inequality become $ \sum {xy(x\plus{}y\plus{}1)}\leq 9$ with $ x\plus{}y\plus{}z\equal{}3$
$ \sum {xy(x\plus{}y\plus{}1)}\leq 9\Leftrightarrow 4(xy\plus{}yz\plus{}zx)\leq 9\plus{}3xyz$ which is schur :wink:

Sorry [b]toanIneq[/b], but I don`t understand why the last inequality is [i]Schur`s inequality[/i].
I know that if $ x,y,z\geq0$, and $ x\plus{}y\plus{}z\equal{}1$, then \[ xy \plus{} yz \plus{} zx \leqslant \frac{{1 \plus{} 9xyz}}
{4} \Rightarrow 4\left( {xy \plus{} yz \plus{} zx} \right) \leqslant 1 \plus{} 9xyz \equal{} 9 \plus{} 3xyz???\]  :|  :maybe:  :!:  :?:

Mîndrilă Claudiu, $ x\plus{}y\plus{}z\equal{}3.$ :wink:

You`re right [b]arqady[/b]. I`m sorry! I`m a bit tired.  :blush:


Putnam
algebra
polynomial
superior algebra
superior algebra theorems

Let $A_i\in M_n(C), i\in\overline{1,n}$ be $n$ nilpotent matrices which commute one with each other. Prove that their product is $O_n$.

There are some problems on the forum where an identity is used (I think Harazi attracted attention on it :)). It concerns expressing the product of $n$ numbers in terms of the $n$'th power's of sums made up from those numbers. I think I know the identity, but I'm not sure right now, and I don't want to make false claims, so I won't state it in general terms. However, I think the pattern is easy to observe (I'll write it for $n=2,3$):

$2ab=(a+b)^2-(a^2+b^2)$ and $6ab=(a+b+c)^3-[(a+b)^3+(b+c)^3+(c+a)^3]+a^3+b^3+c^3$, and so on. The coefficient in front of $a_1a_2\ldots a_n$ is $n!$.

Anyway, we can apply this here exactly as it is because every sum of matrices from the set $\{A_i\}$ is nilpotent, so its $n$'th power is $O_n$. We apply that identity, and we find the product to be $O_n$ as well.

anotherapproachis to notice that the rank $r_k$ of $A_1A_2..A_k$ is strictly decreasing

[quote="jmerry"]There was a problem on this year's Putnam in that vein; you can do it by summing $(a_1\pm a_2\pm a_3\cdots \pm a_n)^n$ over all possible signs, with coefficient (-1)^(# of - signs).

This requires commutativity, of course, and we have that here.[/quote]

I've checked a few cases with Mathematica, and something similar to the Inclusion-Exclusion Principle formula also works, just like the cases for $2$ and $3$ that I wrote in my first post on this topic. I don't expect it to be ahrd to prove, just boring :).

alekk.. and how do you prove that the rank $r_k$ is stricly decreasing?

Let's make another version of Grobber's solution!!
 We have $\left( \sum_{i=1}^{n}A_i\cdot x_i\right)^n = 0$ identically as polynomial
 The coefficent of $x_1\cdot x_2 \dots \cdot x_n$ is $n!\cdot A_1 \cdot A_2 \dots \cdot A_n$ hence....

Kubba, use the fact that $Ker A_i$ is stable by $A_j$ and that the matrices are nilpotent ..

Grobber, proving that identity is not a boring nor easy task. There is a magnificient solution in 3 lines with no computations.

I didn't find this "proof from the book" as Erdos would have called it :) for the identity, but the following is enough for this problem.

We write that formula ($(a_1+\ldots+a_n)^n-\ldots$) in $\mathbb R[s_1,s_2,\ldots,s_n]$, where $s_1=\sum a_i,s_2=\sum a_ia_j$ with $i<j$, and so on (they're the fundamental symmetric polynomials, or whatever they're called :)).

Our polynomial has degree $n$ and becomes null if any of the variables assumes the value $0$, so it's divisible by $s_n=a_1a_2\ldots a_n$, and, since $s_n$ has degree $n$, our polynomial must be a multiple of $s_n$, i.e. it's $t\cdot s_n$ for some real $t$. This suffices for the problem :).

Hope it's correct.

Nice! Here is the solution I knew: consider the product ${ (e^{xx_1}-1})(e^{xx_2}-1)...(e^{xx_n}-1)$. Develop it and use the generating series of the exponential. Identify coefficients and you will have the identity.


AMC
AIME
USA(J)MO
USAMO
ARML
MATHCOUNTS
AIME I

The general consensus about this AIME seems to be that it was significantly easier than previous years' exams. What do you think -- should the AIME be of greater difficulty or not?

My take is that having so many problems of only medium difficulty at best changes what the exam measures. In particular, it changes the focus from the number of problems one can solve to how carefully one works and how few errors one makes. I think that having the former focus is better suited to the AIME and that the latter can be better assessed with the AMC-12.

I think a harder AIME would be better because then it's more about solving difficult problems as opposed to solving a lot of problems and not making careless mistakes. Then again, if its a hard AIME, making one careless mistake could cost you a lot more, too. Anyway, they are just making the tests easy so that 'regular' people can get natural number scores  :P.

Yes, I agree with you Paradox. An easy AIME is probably a detriment to those who are really good at figuring out hard problems (although I do not know this for sure because I am not one of these really good people). It will be interesting to see if there are lower scores on the USAMO because of less qualified people taking the place of more qualified people due to the accessibility of the AIME. Unless the USAMO is easier too...

well..... an easy AIME I may put people taking the AIME II at a disadvantage, IF the AIME II is more difficult (and vice versa). i'm taking the second AIME and praying that both are of the same difficulty, b/c as of now, the cutoff for AIME is projected to be about a 9... and if this second AIME is hard... then the scores will be much lower than those of the 1st AIME. imo, i dont think the difficulty of the tests matter, so long as the first and second ones are equal difficulty. therefore, the cutoff will be adjusted accordingly, and those who deserve to make it will make it.

It really doesn't matter.

AIME has very little relevance for USAMO anyway. They are two completely different tests (in both style and substance).

of course, for the selection of the usa olympic team, the difficulty of aime is insignificant as all the best people will do well basically no matter what.
but for people just hoping to get into usamo, it would be advantageous for the AIME that they're taking to be easier

darktreb has an excellent point. A person who scores highly on the AIME may not obtain a natural number of points on the USAMO. However, a high scorer on the USAMO will generally (not always) score towards the higher end of the AIME scale. It seems that this year these standards have been upset. This AIME may definitely have decreased the average USAMO-level-question ability of its qualifiers. The difficulty of the USAMO itself will also determine how the scores turn out.

Pranay, if you're taking AIME II, you shouldn't worry about how easy or hard it is in comparison to AIME I. The number of qualifiers is supposed to be proportional to the number of students taking either test, so your chances should not change.

[quote="azndsh"]
but for people just hoping to get into usamo, it would be advantageous for the AIME that they're taking to be easier
[/quote]

but then again, basically you are competing against the peopel who took your AIME for USAMO selection becuase there are proportionality between selection from AIME I and II.

[quote="JGeneson"]Where does it say there is a proportional number of qualifiers from AIME I and AIME II? I see that they mention something about a proportional number from both AMCs on the AMC website, but I see nothing about AIMEs. If someone could point out where this is on the site, that would be great.[/quote]

Well, it doesn't exactly say AMC either. "A" or "B" is kind of vague. I've always taken it to mean AIMEs, but I really don't know...

To adjust for variations in contest difficulty, the number of students selected from A & B contests will be proportional to the number of students who took the (A & B) Contests.

they mean by AMC, not AIME. AIME are not lettered a and b, instead they are called The AIME and The Alternate.

You're probably right. But I personally think having an adjustment in the AIMEs is a lot more reasonable than the AMCs. And many people take both AMCs, so how on earth do they make it proportional?

I agree, but it is the AMC's problem, not our concern... although we are affected by it. I think they do take into consideration the proportion of AIME I and II. It just might not say in the Website. You should really consult the experts (hint hint Steve Dunbar).

[quote="JGeneson"] It will be interesting to see if there are lower scores on the USAMO because of less qualified people taking the place of more qualified people due to the accessibility of the AIME. ...[/quote]

I don't agree with "less qualified people".  Why would more qualified people not make it to the USAMO and how would they be replaced with those not as qualified?  I am sure there will be many double digit scores among the more able ones.  The cut-off will be higher.  And there will be some good ones who will miss the mark and some not so good ones who will make it.  That happens anywhere and on any test.

The non-zero scores for the younger kids and the 'not as skilled' ones will be encouraging - at least enough to make them want to come back and try.

Make it overwhelming and you can drive away some young potential candidates.  As a math coach, I see the middle schoolers wanting to come back.  Last year it was different.

[quote="yif man12"]

I totally agree with what you said about middle schoolers wanting to come backin the future. I was in this situation with some of my friends in 7th grade. I doubt we would still be able to take the AIME with any confidence at all in high school or 8th grade if it wasn't such an easy AIME. True, none of us got more than 3 problems correct, but we could solve a few of them, instead of sitting behind a test we were totally stumped by for three hours.
[/quote]but for anybody below the supposedly-grade level suitable for taking AIME and obtaining scores to qualify for USAMO should be hard... that is why anybody in Middle SChool qualifying for USAMO is not only a good thing, but "exceptional". That is why AIME has too be not so easy...

but i still agree with the fact that if it IS too hard, it will discourage some people.

But at the end, those interested will keep trying to work hard, and it will motivate others as well.

Okay, well...I don't think it's great to have an impossibly difficult AIME that has a USAMO floor of 5 or so. But then again, I don't think the difficulty level should necessarily be decreased just for the sake of raising morale among the less experienced qualifiers. Making it to the USAMO has a certain level of prestige to it. Normally, the AIME is difficult enough to test the more advanced problem solving abilities of its takers so that indeed the very best problem solvers advance to the next round (with some leniency towards the lower grades).

A problem arises when students are able to use much less advanced methods (albeit with more time) to obtain answers to problems that normally would require more elevated techniques. A point stands that the best students should still be able to get the questions right, but that's not the issue. It's not only who should be able to get the right answer but also who shouldn't be able to get the right answer. Making more problems more accessible gives away points that would not normally be awarded. Now the question is: if the AIME was USAMO style (aka rigorous solutions only) how many people would've gotten the score they reeled in last week? Or better yet...how about a score in the same percentile?

And now I think we're all getting a little too obsessed about this. But I really shouldn't be one to talk. *nods to self*

Aha, to check my solutions, I wrote a little mini proof for each one.

[quote="JGeneson"]I disagree. AIME is what gets you into USAMO. It might not matter once you are in, but an easy AIME can significantly change the USAMO qualifier pool.[/quote]

Not really, AIME is more about problem solving and USAMO is about knowing how to do proofs-- invariance, graph theory, etc.

My two cents on the matter:

I am against "perfect-down" contests and for "zero-up" contests.  That is to say, I don't like contests where minimizing errors is the largest variable in determining placement, but I do like contests where there are sufficiently many hard problems that it's actually a matter of what you can solve. In particular, I find mundane yet computationally intensive problems disturbing. On the other hand, I like coming across nice problems that I don't know how to solve.

Note that zero-up does not necessarily mean impossible; a zero-up contest could have relatively easy problems as well, the term as I see it merely implies that it is challenging enough for everyone. An example of a contest that is laudably consistent in being zero-up is the USA Computing Olympiad.

Unforunately, for people who are shooting for USAMO and/or high indices (or fortunately if you dissent with my humble opinion), I believe the trend in recent years has been away from zero-up and towards perfect-down.

Great point, Mildorf. I like your idea of a USACO-style math contest. This way, a few silly mistakes would not take someone out of the running. Another good aspect of USACO is multiple contests during the year. If multiple AIMEs decided USAMO, that would leave even less up to chance. What do others think about this idea?

As for what I wrote previously being contested (a few posts up), I do not see anything erroneous with my statement. I simply stated a fact: AIME is the primary test that decides USAMO qualification. If its difficulty is inconsistent, this will result in a somewhat inconsistent set of qualifiers. I did not mean to imply anything else than what was written -- I do agree the AIME and USAMO are significantly different contests. However, the AIME is still relevant to the USAMO.

[quote="mildorf"]
I am against "perfect-down" contests and for "zero-up" contests. That is to say, I don't like contests where minimizing errors is the largest variable in determining placement, but I do like contests where there are sufficiently many hard problems that it's actually a matter of what you can solve. In particular, I find mundane yet computationally intensive problems disturbing. On the other hand, I like coming across nice problems that I don't know how to solve
[/quote]

I agree... that is why i am trying to stay away from Mu Alpha Theta as early as next yr (since ive paid i have to go to this year's national convention, which is in Hawaii... not so bad :) )
and try to do more problem-solving contests like AMC series, ARML, and USAMTS(I forgot to start, so i gave up... bunch of schoolwork and others attributed as well :( )

I'm going to rant a bit in the spirit of the rest of this thread. Earlier, Mildorf mentions a possible different system of qualification for the USAMO. I have to agree with him - a different system will be much better for determining the best students in the nation. Let's take a look at the system as it stands now:

At the base level there are four different exams to qualify one to the next round. Altough they differ in difficulty, they are not very difficult in terms of problem solving. The problems are frequently tricky, the multiple-choice format emphasizes attractive wrong answers, and there is a very short amount of time in which to do work. While this kind of contest works great for say Mathcounts, it simply is not an intelligent way of determining who should advance to the next stage. Right off the bat, the students with speed and precision are given a great advantage (and it seems that in almost every branch of competitive mathematics this is true.) On the other hand, the capable but slow student is disadvantaged.

The next level - the AIME. In the past, a contest of tremendous difficulty, full of challenging problems. However, recently, the trend is toward easier problems that often succumb to the tricks of the experienced problem solver. It also gets harder and harder to determine the top few students. When a score of 12 is achieved by only about 10 people in the nation, it is very easy to determine the top, but if you have over say 100 scores of 12, it is much more difficult. A harder AIME, even one so hard to have a floor of 5 would be much more beneficial. Those who can only solve a limited number of problems have more time to check over their answers, ensuring precision. The importance of stupid mistakes is reduced as well since scores are much more fanned out. The ever-present "bad days" and "test anxiety" is reduced too.

And while the methods of qualification for the USAMO more and more emphasize speed and precision over problem-solving skill and creativity, the USAMO moves away toward a contest where creativity is the greatest tool. The USAMO has expanded toward the level of the IMO, a huge 9-hr exam, while the former AHSME has been cut down on time and difficulty.

Why the exams getting easier at the lower level? To ensure that no one is scared off. But I don't think the level of intimidation was ever a huge problem, especially for AIME-takers. Quite the opposite - with the formerly very consistent AIMEs, it was clear what scores were good. Students knew what to expect, what to try for, and how to evaluate their performances. Nowadays, this is much harder.

So, to conclude this near endless rant, I think that the system is in need of some changes. At the very least, a trend toward more difficult rather than easier exams.

Hmm . . . the goal of selecting USAMO qualifiers is to take the top 250 problem solvers from the AMCs and the AIMEs. The top 250 should not change at all, regardless of the difficulty of the AIME. Just wait and see -- I sincerely doubt that there will be any significant difference in USAMO scores this year. In the meantime, look at past years. Is there a correlation between the difficulty of the AIMEs and USAMO scores? No . . .

I agree with what you all have said about more difficult (or normal, rather) AIMEs being the way to go.

While the AIME seemed easier to AoPSers I do not think the scores overall will be much higher than in previous years. Also, not making any mistakes is a key part in solving a problem. However, I do agree that the format drastically changes from AMC/AIME to USAMO and the difficulty also drastically increases. Thus, the best proof writers do not always qualify for that exam since they tend to be slower workers and might make more careless errors than the "speedy precision workers". Although this seems like a problem, if you think about it this has always been the case. Overall, I do not think the scores this year will be much higher than in previous years although I agree, the tests did seem easier. We must keep in mind that despite the high scores reported on this site, the majority of test takers probably did not feel this way.

A hard AIME is more interesting, but it puzzles me that everyone seems to think this year's was particularly easy. I got a 5 and it didn't seem much different from previous years to me. Still, the harder the better, as long as there are at least a few that should be doable by most qualifiers' standards.

[quote="joml88"]I think that for people who were/are on the borderline of qualifying for USAMO this year probably seemed easier.  For those who were just looking to get a couple right it probably seemed about average.  As for people aiming for perfect I would say it was hard only because there have been no reported perfects yet  :D .[/quote]

Yet many people had the skills to get a perfect. Getting a 14 due to #3 or something like that is probably just as good as a perfect.

I don't like how AIME points were so easily obtained this year. They, after, are part of MOP qualification.

This year was easy, there was like only one hard question(#14), everything else pretty doable(extremely easy #15, #13 can be done easily for anyone who had done AoPS Vol. 2), but still I think the 2003 was just as easy though.

An easy AIME wouldn't be good for people who are aiming for USAMO because it's very easy to make a careless mistake on AIME, or get pressured and panic during AIME, but it's good for beginners who's going to aim for a 2 or 3.

I like easy AIME because i don't get frustrated.

I think that this year's problem was not that It was too easy.  While it was an easy AIME, I think that the more important issue is the consistancy of problems getting more difficult.  In a perfect world, the AIME problems would get consistantly harder from 1 to 15.  This would allow about two for beginners, but still would separate the cream from the milk (Is that a saying?)  It would also, presumably, allow for a perfect score, and would make the cutoff around 7 or 8.

Good point alc, yet because everyone has different strengths and weaknesses, such coordination is nearly impossible.

(The saying you're looking for is probably "cream of the crop".)

I didn't like the easy AIME, but probably for a different reason than everyone else.  As an 8th grader, and this being my first AIME, I don't expect to make it to the USAMO.  However, I do hope to be on the borderline next year.  I thought of this year as a way to get experience, find out what to expect, and to predict my score for next year and years after that.

      The easy AIME did not help this, and with a score of 5, I have no idea whether I will get 2-3, the same score, or even 8-10.

dasherm: There is no way to use one AIME to accurately predict your score for another. It just depends on how you are feeling that day and how your strengths and weaknesses appear on the contest. If you do not believe me, I use myself as an example: My AIME scores have been 7, 2, 10. Will I get a 1 or a 13 next year? There is no way to tell.

That makes me happy!! cuz then life is full of opportunities(like AIME)

Yuck.  Easy AIME = bad.  Too much hanging on not making stupid mistakes.  1-12 were way too easy.  :( Now I can't feel special when I get 11 or something.

[quote="MithsApprentice"]Amen to that. Although, I hope AIME Alt is relatively difficult for fairness sake.[/quote]

For fairness sake? It's not fair for those taking the alternate.

[quote="MithsApprentice"]By "relatively difficult" I meant in relation to AIME I, and in this case relatively means approximately the same.[/quote]

Oh, ok. I  thought you meant much harder than the AIME I.  :blush:

What is  your score in AMC AND AIME, MithsApprentice?


Given a sequence $ a_i \geq 0$, $ i=1..100$, $ a_1 \geq a_2 \geq a_3 \geq...\geq a_{100}$.
Such that: \[ a_1+a_2 \leq 100\]
           \[ a_3+a_4+...+a_{100} \leq 100\]
Find maximum of $ S=\sum_{i=1}^{100}{a_i}^2$
I also have a proof but I don't like it.     :ninja:

Surely not easy!
Could you present your solution? ;)  :)

It's Canada, 2000: [url=http://www.cms.math.ca/Competitions/CMO/exam/cmo2000.html]www.cms.math.ca/Competitions/CMO/exam/cmo2000.html[/url]

A solution appears here: [url=http://www.cms.math.ca/Competitions/CMO/solutions/sol_2000.pdf]www.cms.math.ca/Competitions/CMO/solutions/sol_2000.pdf[/url]


inequalities
induction
combinatorics solved
combinatorics

A scientific conference is attended by k chemists and alchemists, of whom the chemists are in the majority. When asked a question, a chemist will always tell the truth, while an alchemist may tell the truth or lie. A visiting mathematician has the task of finding out which of the k members of the conference are chemists and which are alchemists. He must do this by choosing a member of the conference and asking him: "Which is So-and So, a chemist or an alchemist?" In particular, he can ask a member: "Which are you, a chemist or an alchemist?" Show that the mathematician can accomplish his investigation by asking

a.) 4k questions
b.) 2k-2 questions
c.) 2k-3 questions
d.) 3/2*k - 1 questions.

[i]Moscow Olympiad 1979[/i]
http://www.mccme.ru/olympiads/mmo/index.htm

It has been shown that the minimum number of questions to accomplish the investigation is N(k) = [3(k-1)/2] for k  \geq 3 (where [.] denotes the integer part).

The part N(k)  \leq [3(k-1)/2] is not very difficult, and I will give a solution if needed.
The reverse inequality is not too difficult but lengthy...

Reference :
P.M.Blecher, 'On a logical problem', Discrete-Mathematics, 43 (1983), 107-110.

Pierre.

Ok, let's prove that N(k)  \leq  [3(k-1)/2] following the above reference.

- First let's suppose that k = 2p+1 is odd. We want to prove that N(k)  \leq  3p.
The result is obvious for k = 1, since there is only one person which has to be a chemist.
Now suppose the result proved for each i < p.

Let P_1,...,P_N be the persons.
We start to ask P_2,P_3,... if P_1 is a chemist ans stop  as soon as one of the folowing two events will occur :
- Event A : p persons have said that P_1 is a chemist. Then, P_1 is indeed a chemist and all those who said 'no' are alchemists (if P_1 was not a chemist then among the 2p others persons there are a majority of chemists and they all tell the truth, so there can not be p answers 'yes'). We ask now the first about all the others persons. Direct computation shows that we use 3p questions in this procedure.

- Event B : The number of those who said 'no' exceeds the number of those who said 'yes'. In that case, if m persons said 'no' then m-1 persons said 'yes', m = 1,2,.... Moreover, it is easy to verify that in the group of the first 2m persons the number of alchemists is not less than the number of the chemists, so in the rest group of 2p+1 - 2m persons the chemists form a majority. So using 3(p-m) questions, we can sort out them (from the induction hypothesis). Next, we choose a chemist from them and ask him about P_1 and about those among the first 2m-1 persons whose answers about P_1 were truthful (others are evidently alchemists). Thus, we sort out all the persons by no more than 2m-1 + 3(p-m) + 1 + m = 3p questions.

The case where k is even is similar.

And we are done.
Pierre.


Solve the follwing equation with respect to $x$.

\[\frac{(x^2-x+1)^3}{(x^2-x)^2}=\frac{(a^2-a+1)^3}{(a^2-a)^2}\]

[quote="amd2357"][hide]Set
\[\displaystyle f(x)=\frac{(x^2-x+1)^3}{(x^2-x)^2}\]
After trying a few transformations we find
\[\displaystyle f(x)=f(\frac{1}{x})=f(1-x)\]
Since $x=a$ is obviously a solution we have $\displaystyle x=\frac{1}{a}$ and $x=1-a$ as solutions.  Additionally $\displaystyle x=\frac{1}{a}$ implies $\displaystyle x=1-\frac{1}{a}=\frac{a-1}{a}$ is a solution and $x=1-a$ implies $\displaystyle x=\frac{1}{1-a}$ is a solution.  Finally, $\displaystyle x=\frac{a-1}{a}$ implies $\displaystyle x=\frac{a}{a-1}$ is a solution giving us all six roots.  I don't know of a more direct way to find the necessary transformations.  Any ideas?[/hide][/quote]

You are right,amd2357. :)

Judging from the form of the answer, this problem is related to Mobius Transform, I think.

kunny


Let $ ABC$ be a triangle ($ \hat{A} \equal{} 90^o$). Draw $ Ax \parallel BC$. Let $ D \in Ax$ such that $ BD \equal{} BA$. Draw $ AH \perp BD$. Prove that $ CD$ meets $ AH$ at the midpoint of $ AH$.

Edit: $ AH \perp BD$

NO. IT'S TRUE.
Can you tell me the reason why you said it's not correct, dvtvd?

Sorry :blush:  :blush:
I have edited the problem: $ AH \perp BD$

Now the problem is true, can someone help me ?  :(

Nice problem!
We know $ \angle BDA \equal{} \angle DAB \equal{} \angle B$. Let CD intersect AB at X.
By Menelaus, we have to show $ \frac {DH}{DB} \cdot \frac {BX}{XA} \equal{} 1$.
We know $ DH \equal{} DA \cdot cos B$ and $ DB \equal{} \frac {DA}{2 cos B}, so \frac {DH}{DB} \equal{} 2cos^2B$
Also $ \frac {AX}{XB} \equal{} \frac {[ADC]}{[DBC]} \equal{} \frac {AD \cdot AC sin(90 \plus{} B)}{BD \cdot BC sin (180 \minus{} B)}$. Write everything in terms of AB, this cancels to $ 2cos^2B$, which finishes the proof.

My method is messy, but I believe there is a nice proof by taking the midpoint of a side and using the orthocenter (haven't found it, just have a feeling).

I'd like to post my solution, which I found after a lot of time.Correct me if I am mistaken :|

Suppose that CD meets AH at M. We draw the circle (B,BD), which will be refered as C1. If E is the second point that DH meets C1 we have:<EBC=<EDA
<CBA=<BAD, <BAD=<EDA. Clearly, the triangles EBC,CAB are equal. Thus, as CA is tangent to C1, CE is also tangent to C1. So CB is perpedincular to AE and CB meets AE in its midpoint K. If CD C1 again at P we have: CA^2=CP*CD
CA^2=CK*CB. So:CP*CD=CK*CB and P,B,K,D lie on the same circle. So <KPD=<EBC=<ABC=<ADB. <HAE=90-<HEA=<ADB, as DE is a diameter of of C1. So <KMP=<KAP=<EDP and KM is obviously parallel to ED. Thus, KM//EH and as K is the midpoint of AE, the proof is completed. :!:

panos_lo, your solution is very nice but it'll be nicer if you use $ latex$
And thank you dgreenb801 too

Let $ M \equiv AH \cap DC$ and $ N \equiv AB \cap DC.$ By Menelaus' theorem for $\triangle AHB$ cut by line $\overline{DMN},$ we get

$\frac {AM}{MH} = \frac {AB}{DH} \cdot \frac {NA}{BN}$

But since $ \triangle ADN \sim \triangle BCN$ and $ \triangle ADH \sim \triangle CBA,$ we obtain

$ \frac {NA}{BN} = \frac {AD}{BC} = \frac {DH}{AB} \Longrightarrow  \frac {AM}{MH} = \frac {AB}{DH} \cdot \frac {DH}{AB} = 1  \Longrightarrow  AM = MH.$

luis, thanks for your very nice solution :lol:


Does anyone know when the AIME I results will be out?  I know they are emailed to the teacher, but our school is on spring break.  I was wondering if you could find the list of USAMO qualifiers on the UNL website.

yes, a list of qualifers will be listed soon...I except on or before next week...

The AMC office mailed the AIME I results and reports from Lincoln last Friday, April 2.  They should be reaching schools by now.  Check with your teacher or contest manager.  When the AIME II results will be mailed depends on how soon the AIME II scoresheets reach the AMC office.  Schools are supposed to "overnight" the scoresheets to the AMC office, and a few have already arrived, but until all scoresheets are received, we aren't done and can't compile statistics and reports.

Yes, the names of the invited USAMO participants are posted at the AMC website, but not until we we have a complete list.  In turn that depends on getting all of the AIME II scoresheets in, and in turn that depends on the schools sending the results in a timely fashion.

Steve Dunbar
AMC Director


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Please, contact the webmaster. Thank you.

What were you doing when this error happened?

I was going to page 2 of the either Interm or Intro Counting archive; I can't remember which, and the second time I tried it worked.

That happened to me a loong time ago in interm counting archive but I just refreshed and it worked.


An expression is formed using the numbers 7, 16, 25, and 27 according to the following rules.
Each of the four numbers is used exactly once. \\
The four numbers may be used in any order. \\
Exactly three operations are used; each one is either $ \plus{}$ or $ \times$ \\
An unlimited number of parentheses may be used. \\
No two distinct expressions have the same simplified value. The two expressions below are not distinct, and therefore must be counted as only one value. What is the greatest number of distinct values, including the one below, that can be obtained when building expressions following these rules?

[asy]draw((0,0)--(2,0));
draw((3,1)--(4,1)--(4,2)--(3,2)--(3,1)--cycle,linewidth(1));
draw((5,0)--(7,0));
draw((8,1)--(9,1)--(9,2)--(8,2)--(8,1)--cycle,linewidth(1));
draw((10,0)--(12,0));
draw((13,1)--(14,1)--(14,2)--(13,2)--(13,1)--cycle,linewidth(1));
draw((15,0)--(17,0));

draw((18,1.5)--(20,1.5),Arrow);[/asy]
$ (7\plus{}16\plus{}27) \times 25 \equal{} 25 \times (27\plus{}7\plus{}16)$

[hide="Solution"]Casework using the number of addition and mutliplication signs:

(i) [i]3 addition[/i]
$ a \plus{} b \plus{} c \plus{} d$ -> $ 1$ way

(ii) [i]2 addition, 1 multiplication[/i]
$ (a\times{b}) \plus{} c \plus{} d$ -> $ \binom42 \equal{} 6$ ways
$ a\times{(b \plus{} c)} \plus{} d$ -> $ \binom42\times{2} \equal{} 12$ ways
$ a\times{(b \plus{} c \plus{} d)}$ -> $ \binom41 \equal{} 4$ ways
$ (a \plus{} b)\times{(c \plus{} d)}$ -> $ \binom42\div{2} \equal{} 3$ ways

(iii) [i]1 addition, 2 multiplication[/i]

$ (a\times{b}\times{c}) \plus{} d$ -> $ \binom41 \equal{} 4$ ways
$ a\times{b}\times{(c \plus{} d)}$ -> $ \binom42 \equal{} 6$ ways
$ a\times{(b\times{c} \plus{} d)}$ -> $ \binom42\times{2} \equal{} 12$ ways
$ (a\times{b}) \plus{} (c\times{d})$ -> $ \binom42\div{2} \equal{} 3$ ways

(iv) [i]3 multiplication[/i]

$ a\times{b}\times{c}\times{d}$ -> $ 1$ way

Adding these up gives a total of $ \boxed{52}$ ways.[/hide]

Is there a solution besides massive casework?

This was a National Target Round Problem #8.

I do not believe there is another solution. 'RunpengFAILS's was the best one.  :)

the answer is the 5th bell number, so i think this would be a more elegant solution than just massive casework as other people answered

there are 4 numbers and then a choice of opperations so this could be seen as how many ways are there to partition a 5 element set, or B5 of the bell numbers

http://mathworld.wolfram.com/BellNumber.html

[quote="H.T.W.U."]the answer is the 5th bell number, so i think this would be a more elegant solution than just massive casework as other people answered

there are 4 numbers and then a choice of operations so this could be seen as how many ways are there to partition a 5 element set, or B5 of the bell numbers

http://mathworld.wolfram.com/BellNumber.html[/quote]
I'm pretty sure it is just a coincidence that 52 is a Bell Number.

If you consider 3 numbers, and ask for the maximum number of distinct possible answers using the operators "plus", "times", and unlimited parentheses, I count 8 distinct results.  But 8 is not a Bell Number.

The answer by RunpengFAILS looks pretty good to me.  [You might manage with half the casework, if you notice the casework always comes in pairs, through interchange of all + and * signs with each other.]

Not quite sure: what is a Bell Number?

Also, when do you know when to use casework or complementary counting...?

[quote=arthurwms]An assumption in both the official solution and this thread is that the only duplicates are due to the commutativity and associativity of the operators + and x themselves, so you can use a, b, c, d instead of the given numbers 7, 16, 25, 27 to get the same number 52 of distinct expressions, but that's not stated in the problem.  Is it implied by some simple property of the set { 7, 16, 25, 27 }?  It's certainly not true in general, for example with the set { 1, 2, 3, 4 } there are 21 distinct solutions, and for the set { 2, 3, 5, 7 } there are 44 distinct solutions.[/quote]

Yes this is bugging me too. How can the problem writers ensure that an expression formed using $7,16,25,$ and $27$ does not have any duplicates just by coincidence. For example, in $S\in {1,2,3,4}$, we have that $3\times 2+4\times 1=10=1+2+3+4$; using the counting strategies in @RunpengFAIL's and alcumus's sol we find that the two expressions are distinct although their values are the same. Is this because $1,2,3,$ and $4$ are consecutive integers?

This problem was so annoying and that’s why I have abandoned it for a long time. I came again today and spent like almost an half-hour seeing if there is a simpler solution and resolved to a massive casework at the end. Lmao im glad it’s over.


In a 120 × 150 rectangle (made out of unit squares joined along their sides), how
many unit squares does its diagonal pass through?

[hide="Hint"]PIE:
How many horizontal lines does the diagonal cross?
How many vertical lines does the diagonal cross?
How many intersections does the diagonal cross? [hide="Hint for the last question"]What information gives you $ gcd(120,150)$?[/hide][/hide]


[quote="akalra1"]yea seriously give us a LINK[/quote]

seriously stop being lazy and GOOGLE


inequalities
geometry
circumcircle
trigonometry
geometry proposed

Let $D$, $E$, $F$ be points on the sides $BC$, $CA$, $AB$ respectively of a triangle $ABC$ (distinct from the vertices). If the quadrilateral $AFDE$ is  cyclic, prove that \[ \frac{ 4 \mathcal A[DEF] }{\mathcal A[ABC] } \leq \left( \frac{EF}{AD} \right)^2 . \]
[i]Greece[/i]

[code]\frac{ 4 \mathcal A[DEF] }{\mathcal A[ABC] } \leq \left( \frac{EF}{AD} \right)^2 .[/code]

[color=purple]What does $ A[DEF]$ mean ? [/color]

[code]\mathcal A[/code] is area of appropriate triangle

Let $ O$ be the circumcircle of $ ABC$ ,$ AD \cap O \equal{} K$
It's easy to see that , $ \triangle{DEF} \sim \triangle{KCB}$
so $ \frac{[\triangle{DEF}]}{[\triangle{KCB}]} \equal{} \frac{EF^2}{BC^2}$ , $ \frac{[\triangle{KCB}]}{[\triangle{ABC}]} \equal{} \frac{DK}{AD}$
thus $ \frac{[\triangle{DEF}]}{[\triangle{ABC}]} \equal{} \frac{EF^2}{BC^2}\cdot \frac{DK}{AD}$
It's remain to prove that $ 4DK \cdot AD \le BC^2 \leftrightarrow 4BD \cdot DC \le BC^2$ which is obvious because $ BC \equal{} BD\plus{}DC$

Very nice DJMATHMAN!

[quote="djmathman"]I have an alternate solution.  (Sorry about all the edits; I originally had an error in this solution, but then I fixed it.)

Let $\angle BAD=\alpha$ and $\angle CAD=\beta$.  Note that the inequality can be rewritten as \begin{align*}[ABC]&\geq4[DEF]\left(\dfrac{AD}{EF}\right)^2\\&=2\left(\dfrac{DE}{EF}\right)\left(\dfrac{DF}{EF}\right)AD^2\sin(\pi-\angle A)\\&=2\left(\dfrac{\sin\alpha}{\sin\angle A}\right)\left(\dfrac{\sin\beta}{\sin\angle A}\right)AD^2\sin\angle A\\&=\dfrac{2\sin\alpha\sin\beta\cdot AD^2}{\sin\angle A}\end{align*}  Therefore it suffices to prove that $[ABC]\geq\dfrac{2\sin\alpha\sin\beta\cdot AD^2}{\sin\angle A}.$

Now we examine $\triangle ABC$.  Note that $[ABD]=\dfrac12(AB)(AD)\sin\alpha$ and $[ADC]=\dfrac12(AD)(AC)\sin\beta$, so $[ABC]=\dfrac12AD(AB\sin\alpha+AC\sin\beta)$.  But by AM-GM \begin{align*}\dfrac12AD(AB\sin\alpha+AC\sin\beta)&\geq AD\sqrt{AB\cdot AC\sin\alpha\sin\beta}\\&=AD\sqrt{\dfrac{2[ABC]\sin\alpha\sin\beta}{\sin\angle A}}.\end{align*}  Therefore \[[ABC]\geq AD\sqrt{\dfrac{2[ABC]\sin\alpha\sin\beta}{\sin\angle A}}\implies [ABC]\geq\dfrac{2AD^2\sin\alpha\sin\beta}{\sin\angle A},\]  as desired.  Equality holds when $AB\sin\alpha=AC\sin\beta$, which occurs when $D$ is the midpoint of $BC$.[/quote]


Martin is really bad at adding fractions.  Instead of finding a common denominator he just adds the numerators to get the
new numerator and adds the denominators to get the new
denominator.  He tries to add $\frac{2}{3}$ to some simplified fraction only to get an answer of  $\frac{1}{3}$ .  If the numerator and denominator of the original fraction have the largest sum less than 200 of all such fractions for which he would get this incorrect answer, what is this original fraction?  Express your answer as a common fraction.(duh)

[hide]So we have: (2+x)/(3+y) = 1/3. Cross multiply to get 3+y = 6+3x. y = 3x+3
We also have x+y<200. Plug 3x+3 in for y: 4x+3<200 and the largest integer x can be is 49. Therefore, y = 3*49 + 3 = 3(49+1) = 150.

Just to check, 150 + 49 = 199 and (2+49)/(150+3) = 51/153 = 1/3.

The answer is [b]49/150.[/b][/hide]


What do you think of the field $\mathbb{R}$? Do you agree, that it is extremely ugly?

I don't get it. Why would the reals be ugly??  :|

I can kind of see (vaguely) what is meant by "ugly", although I can't think of any concrete example of this as applied to a field.

In another area, however, I would say that the factorial function is an example of something ugly, mainly because of the way it doesn't behave well with other functions (nothing even remotely resembling a distributive property, for example).

As for the reals specifically, I think the fact that R is totally ordered makes it pretty.

Hmm, strange; I didn't expect so many real-loving guys around here...
$\mathbb{R}$ has almost no nice algebraic properties. For example it is not algebraicly closed, as for example $\mathbb{C}$. You might argue, that $\mathbb{Q}$ or $\mathbb{F}_p$ are not closed either, but hey! they are prime fields which is quite nice and surely you wouldn't expect from prime fields to be algebraicly closed!
But even calculus which seems to have the closest relationship to $\mathbb{R}$ , one often prefers to work over $\mathbb{C}$.
But I'm not convinced by your argument that $\mathbb{R}$ is pretty, because it's totally ordered; this applies to $\mathbb{Q}$ as well

[quote="Sly Si"]I guess I didn't mean to say that being totally ordered alone makes R pretty, it's just one reason not to think it's ugly.

In fact, when I said totally ordered I meant something else.  There's some other property of R that no other field has, I believe it's that R is the only totally ordered field in which every subset has a least upper bound and a greatest lower bound.  (What's that property called, by the way?)

Good point about R not being algebraically complete.  It does, though, have a nice finite extension (namely C) that is, if that counts for anything.[/quote]

Completeness Axiom

In fact, (copied word-for-word from one of my textbooks - hopefully I won't get sued):
[quote="Browder, Andrew"][i] We say that $K$ is a[/i] complete ordered field [i]if $K$ is an ordered field with the property: given any nonempty subset $S$ of $K$, if the set $\mathcal{M}_S$ of upper bounds of the set $S$ is nonempty, then $\mathcal{M}_S$ possesses a least element.[/i]

[i]There exists a complete ordered field $\mathbb{R}$. If $R_1$ and $R_2$ are complete ordered fields, there exists an isomorphism of ordered fields between them.[/i][/quote]