AMC
AIME
USA(J)MO
USAMO
geometry
trigonometry
AMC 12

Comments welcome.  Also please comment if you think the problems on one were "cooler" or "cuter" or nicer than the problems on the other.

Much harder. Especially the middle problems.

do people think there will be a different index/floor cutoff for the II?

way harder, but I still am flabbergasted at my 4.

i felt that aime ii was trivial compared to aime i in general. the first 12 were of the same easiness, and then the difficulty jmumped up. in aime i (i only looked at it, that could be why), the difficulty increased exponentially

I think everyone got roughly around the same scores as the AIME I takers, so there shouldn't be different indices.

I wouldn't say this AIME is harder than AIME I, because I got a 10 on this but I got a 5-6 on AIME I while practicing.

I personally think it is about the same level of difficulty. AMC flips coin to decide which one goes to which AIME. so I guess it depends on what type of problem you favor. I took AIME I and got 9. But for AIME II practice, I got 10, perhaps it is less pressure so I made fewer calculation mistakes in practice.

Good luck to all!!!!!!!

I looked at the problems for the AIME II, and it seems a lot harder than the AIME I.  Of course, I haven't actually sat down and attempted to work the problems, so this is obviously biased.

Maybe there is a chance for a 150/6 now?

there probably is if you took the aime i. it looked pretty hard to me. methinks both aimes did a great job of making some easy problems seem intimidating.

I kinda think this is harder too. I took this test unofficially and got a 7. (Got #1-5, 8, and 10 correct) I mostly think that the middle problems are much harder than the AIME I's middle problems...but even though I only got a 6 officially on the AIME I, I still think the II is harder because I made so many stupid mistakes on the I (in addition to wasting time such that I should have spent it doing one of my unsolved problems, which was trivial), therefore it meant I should have gotten something higher on the AIME I...Possibly 2-4 more problems right if it weren't for stupid mistakes...Meh.

The middle problems really slowed me down as I was doing the test and prevented me from having enough time to solve all the problems except 12 [b]and[/b] check my answers.

Compared to previous AIME's and also in general to other math competitions I have done, I think this AIME had a very large amount of brute force/plug and chug involved, with relatively less insight required. This may be because I really wanted to make USAMO this year, and on most problems I went with the first solution method I came up with instead of thinking about it and trying to find a "nice" method.

To me, AIME II was definitely harder than AIME I.

I got 8 on AIME I (even 10), but I only got 5 on AIME II(practice). :oops:

I could solve 11 problems legitimately on I, but I could solve only 7 problems legitimately on II.

I didn't actually try to solve a lot of problems on AIME II, but still...

Hmmm...I thought that AIME II was a lot easier than AIME I

(that could just be because I'm horrible at problems that require a lot of computation, though...)

Honestly, I don't see how the AIME II was significantly harder than the AIME I. The middle problems (6-10) are what typically differentiates between an USAMO qualifier and those who don't, and I personally felt that the middle problems on the AIME I (especially 9 and 10) were much harder than those on the AIME II. On the AIME II, #6 was a pretty generic combo problem. #7 is also a pretty straightforward problem. #8, if you had any experience with a game/series problem (Mathcounts comes to mind), this should have been easy. #9 could have passed for a #2 or 3 in my opinion. And #10... was pretty hard in terms of understanding the problem.

I got an 8 on the AIME I (3 careless) as practice and a 10 on the AIME II (1 careless) officially, so I'd say they're nearly about the same. However, looking at polls on AOPS, it does seem that the AIME II seems statistically harder. Who knows? Guess we'll find out soon enough.

I got 7 for both AIMEs (I official) but I made 2 mistakes on I and none for II. So I guess for me, it seems like I is easier but more error-prone. Of course, I didn't look at "connection stupidity"(failing to draw obvious lines, etc) which I had lots of on both, since it's hard to draw the border between "should have" and "could have" sometimes.

[quote="xpmath"][quote="asdf4"] On the AIME II, #6 was a pretty generic combo problem. #7 is also a pretty straightforward problem. #8, if you had any experience with a game/series problem (Mathcounts comes to mind), this should have been easy. #9 could have passed for a #2 or 3 in my opinion. And #10... was pretty hard in terms of understanding the problem.

[/quote]

What? I thought all of 6-9 on AIME II were harder than their corresponding problems in AIME I.
#10 was certainly easier on AIME II. #6 is basically a generic combo problem, but #6 on the AIME I was pretty easy to see. Same thing for #7.

#8 was actually hard for me... on #8 on the AIME I, I immediately knew what needed to be summed, but I suppose the calculations may have been difficult on that.

I thought #9 on AIME I was one of the easiest #9s I had ever seen. I suppose it doesn't matter what we think; we'll find out soon anyway.[/quote]

What happened to all #9s are easy? :P

That was Temperal. I usually never get #9s.

[quote="xpmath"]
I thought #9 on AIME I was one of the easiest #9s I had ever seen. I suppose it doesn't matter what we think; we'll find out soon anyway.[/quote]
Hehe... I didn't get #9 on the AIME I whereas #9 on the AIME II was probably the easiest #9 I've ever seen. I'm not very good at combinatorics which explains how #9 and 10 were hard for me on the AIME I. Still though, I got #6 on the AIME II. I didn't get #6 on the AIME I either, didn't understand the problem. Both #10s on I and II were hard in my opinion.

I can see where your reasoning goes. I guess if you're better at combinatorics and counting, you'd be better off with the AIME I, since you're more likely to get #8-10. If you're not, then AIME II might've been better. But in our (me and my schoolmates) case, AIME II definitely seemed easier.

Dude, #7 and #9 on II were like really really hard for me, I didn't solve either.

#6 and #9 on the II were pretty hard for me too...didn't get either...though I didn't spend much time on #9...#7 needed a little time to figure out the trick and #8 was pretty straightforward for me though...

In general, I believe it is kind of tough for people to decide, for we could only take one of them. I looked at the other test and worked some, but still, we didn't take both in a testing environment. BTW- I PHAILED

The 2009 [url=http://www.unl.edu/amc/e-exams/e7-aime/e7-1-aimearchive/2009-aa/09-AIMEstats.shtml]AIME Statistics[/url] are up.

Attached are some more in depth grand statistics.

[quote="AMCDirector"]Preliminary List - April 10, 2009
The descriptive statistics for the AIME I and the AIME II are not significantly different, particularly at the upper end of the distribution. The statistics do not indicate separate consideration of the resulting indices.[/quote]

About 4/5 as many people took the II as took the I (4400 compared to 5523).  Compare these numbers:

(score, AIME I scores >= score, AIME II scores >= score)
(15,                     4,                                  1)
(14,                     20,                                4)
(13,                     40,                                12)
(12,                     74,                                31)

These statistics seem extremely different at the upper end, at least to me.

By the time you get down to 8 though, the statistics are more similar, so USAMO qualification seems correct...

In fact the mean, median, and mode for the II were higher than the I.

Looking at the score distribution, it was easier to get a 3, 4, 5, 6, or 7 on the II but harder to get above that.

Looking at the cumulative distributions yields some insight here:
[code]
Score    %AIME I    %AIME II
>=15       0.07%       0.02%
>=14       0.36%       0.10%
>=13       0.73%       0.24%
>=12       1.37%       0.68%
>=11       2.42%       1.45%
>=10       4.92%       3.25%
>=9        7.87%       5.84%
>=8       12.19%      10.23%
>=7       17.78%      16.87%
>=6       25.94%      26.23%
>=5       36.10%      37.94%
>=4       48.32%      54.42%
>=3       63.35%      72.24%
>=2       78.71%      87.81%
>=1       92.06%      97.70%
>=0      100.00%     100.00%
[/code]
In general, I would think that you want scores in the 7-10 range to map over the most consistently, since these seem to be the borderline cases for USAMO selection.  [b]In each of these cases, it appears that AIME I was easier, but not by a full point [/b](perhaps more like 1/2 a point -- a bit less than that around 7 and a bit more than that around 10).

This type of analysis does assume an equal distribution of talent across the two test dates, which may not be true.  Perhaps a complementary analysis would be to fit AMC 12 scores vs. AIME scores across the 4 combinations and look for a shift.  Again, it would make sense only to consider the subset of students with AIME scores near the USAMO cutoff (i.e. the 7-10 range).

The mean, median, and mode are not really interesting IMHO, as these scores are not near the cut line for the USAMO.  If I had to guess, the 931 vs. 964 on AIME I Q1 was responsible for shift at this end of the distribution.

P.S. Here is an attempt at an equivalent score calculation (again, assuming equal talent pools):

[code]
Score  AIME I %   AIME II %    Equiv AIME I score
-----  --------   ---------    ------------------
>=7      17.78%      16.87%            7.16
>=8      12.19%      10.23%            8.45
>=9       7.87%       5.84%            9.69
>=10      4.92%       3.25%           10.67
>=11      2.42%       1.45%           11.92
[/code]
The equivalent scores are simply obtained by interpolating the cumulative AIME II percentages into the AIME I pareto.  Again, it is not a full point for any of these cases, so I think it is quite understandable why no adjustment was made.

I personally thought the AIME I was easier than usual, but I haven't done the II yet (practice).

-rts2007


:D For those chem enthusiasts, pls check out the gen discussion forum for my chem question and give an answer pls.

do not  :spam:
i've already answered it  b4 u posted dis :)

btw
[hide]i mite b the only 1 on aops who likes chem[/hide]

Hey sagar, many people here like chemistry. Yes u gave an answer, but that does not mean others need not see it.The post is for those who do not see gen discussion but instead only olympiad section.

[hide][u][b]@Sagar[/b][/u]: Sorry for that Nitrous oxide thing :rotfl: , i was thinking of NO.[/hide]

oh no don't misunderstand me  :(
i wasn't boasting.it's just that many longtime aopsians like nuclear_alchemist,rajiv,conjurer,rituraj etc (i don't remember who else) have posted in some thread or the other that they hate chemistry, that's all  :D


In a circle we have fixed $ n$ distinct points. Let in each point be a real number : $ x_{1},x_{2},...,x_{n}$,such that $ \sum_{}^{} x_{i} > 0$.
Let's make a game with these numbers. If does exist a negative number,(example $ x_{k}$)then the player makes a step taking that number, adding $ x_{k}$ its nearby numbers , and changes the sign of selected number .Game is finished when all numbers are positive. Prove that this game will end after a finite number of steps!

This one seems to be very popular recently.  It's IMO 1986 problem 3.  I found it by searching for "finite number of steps".

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60773
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=294347
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=265620


quadratics
algebra
polynomial
calculus
integration
complex numbers
superior algebra

I'm not really sure how to tackle these problems when the degree is greater than 3 (i.e. I don't just have to show there are no roots) or I cannot use Eisenstein's criterion.

1.  Show that x^4 + 8 in Q[x] is irreducible over Q.

I determined that all 4 roots were complex but I'm not sure how to show that there are no quadratic terms.  I started the method of undetermined coefficients but it seemed like I would have to go through about 16 cases.

2.  Determine whether or not x^4 - 3x^2 +9 is irreducible over Q.

With this problem, I basically just 'guessed' the factorization -- x^4 - 3x^2 +9 = (x^2 - 3x +3) (x^2+3x + 3).  Is there a more 'normal' route to go beyond guessing when you're demonstrating that this polynomial is reducible?

Thanks,
V

[quote="ZetaX"]1. Note that when using $x=2y$, the polynomial gets (after neglecting rational factors) $2y^{4}+1$. Now any factorisation of the original polynomial gives one for this new one. Set $y=\frac{1}{z}$, then the last polynomial factors give polynomial factors of $z^{4}\left( \frac{2}{z^{4}}+1 \right) = z^{4}+2$. But that doesn't have factors by Eisenstein.
Another way: Any factor must (wlog) have integral coefficients. Especially, the constant term must be integral. Now note that uo to sign this is the product of the roots, so we would have $\sqrt[4]{8}^{k}$ integral for one of the factors, $k$ being its degree. But that happen first only for $k=4$, so the factor has degree $4$.

2. You could substitute $y=x^{2}$, find the roots and re-combine them to get rational polynomial factors.
But better: when $x$ is a root, then also $-x$ is so. Thus when factoring with some irreducible factor being $q(x)$, then $q(-x)$ is also one (only the same when $q(x)=r(x^{2})$ for some $r$). Now all factorings possible are of type $(x^{2}+ax+b)(x^{2}-ax+b)$, which immediatly gives $b=3$ and $a= \pm 3$.[/quote]

Thanks, I really appreciate the explanation.  For #1 ('Another way') -- is there a theorem for that?  Also, I'm embarrassed to even mention it but for #2, I did find the roots but I don't know how to re-combine them to get rational polynomial factors.  We just never did this.  By any chance do you have any references for something like that?  I did understand the other method for #2.

Again, thank you.

V

ad 1: It's an application of Gauss's (Polynomial) Theorem: A polynomial with integer coefficients is irreducible over the rationals iff it's irreducible over the integers.

ad 2: If $\overline{a}$ is the complex conjugate to $a$, then $(x-a)(x-\overline{a})$ is a polynomial with real coefficients. They are even rational when $a \in \mathbb{Q}(i)$. The roots of the polynomial $x^{4}-3x^{2}+9$ have the form $a,\overline{a},b,\overline{b}\in \mathbb{Q}(i)$, so that it is the product of the rational polynomials $(x-a)(x-\overline{a})$ and $(x-b)(x-\overline{b})$. The resulting factorization is $(x^{2}+3x+3)*(x^{2}-3x+3)$.


Some schools don't.  I haven't timed myself in a long time.  But I can say somewhere between 6 and 8 minutes.  I do run track, and my main event is the 400 (probably 600 this year), but surprisingly, I've never timed a mile since 7th grade.

Around 7:00, though I have broken it before. I think that was a timer error though  :wink:

[quote="junggi"][quote="mtms5467"]Some schools run the mile every fall. What's your mile time?[/quote]
some school make us run the mile every quarter. some particularly evil teachers make them run every couple weeks(thank god my teacher wasnt like that).

I ran the mile today, and got 9:46(to me this is actually good. I never ran the mile faster than 10 min for the past year or so)[/quote]
I run around 3 miles every day...today I did this trail called "white loop"...4.5 miles of hills... :wink:

i run a mile every day pretty much... i believe my time is slightly under 7 minutes, never bothered timing it

[quote="mtms5467"]Some schools run the mile every fall. What's your mile time?[/quote]

i'm breaking into 7 minutes, probably because my gym teacher makes us run around the track daily for a warmup for whatever torture she has planned for us. what's your best time for the 100 meter dash?

My horribly cruel gym teacher makes us run a mile a day first period to "warm up". I think that I'm around 7 minutes though.

My best ever was 6:51 but I haven't practiced much lately and I think I'm about 30 seconds slower.

My best is 6:50 (last year), and I think I can do better if I try. :D

6:23, but I'm pretty sure that was a timing mistake and it was actually 7:23. That was last year though.

We haven't run it this year yet. My best time for the mile is 7:34, it's not that bad considering I'm a girl.

I do XC (aka CC) and track.

I'm not taking PE this year, but my best time was probably 8:37 :roll: And even that was probably a timer error. :pilot:

You know you're a math nerd if it takes you four times longer to run the mile than it does to solve Rubik's Cube...

i have asthma too, mainly triggered from the cold and dust.  my best time ever was a 8:45

well strangely enough, i think i have some form of exercise-induced asthma.

that may be what many of you guys are talking about. its fairly common but many people are oblivious to the fact that they have it. if you are fine almost all the time, except after long runs, then that is a sign. i would check with a doctor next time. not to big of a deal because there are inhaler things that would keep down a lot of coughing you get from running.

nah well kinda but not only excercise(which i think is a no)

I walk like half the mile, but I don't have asthma. :D

last time I timed was a few years ago and that was 8:30, probably below 8 now  :D

[quote="me@home"]This morning I ran 9 miles of hilly stuff at a local state park with my XC team... I'm not that good on cross country, just good enough to be last on varsity/ first on JV in most meets.
[quote="drunner2007"]i have also run the 2 mile and 5k. here are my best times

1mile: 4:49
2mile:10:32
5k:16:05

by the way, that's awesome that you made the MIT team joml88.
sounds like a lot of fun[/quote]
Thats wierd, you should be able to do much better at 1 and 2 mile (unless those are your splits)
a 16:05 is pretty fast, my pr this season is only 19:05... (but I made the local paper for it because my coach liked my negative splits so much)
and I don't really do track (ran like 3 meets) so out of those my mile time is a sluggish 5:02...[/quote]

You guys 2 mi. scare me.

Don't be scared of 2 mi, the faster you go the shorter it takes...

[quote="K81o7"]I hear you have to break 4:20 to make it onto the MIT cross country team  :what?: ...is that true?[/quote]

what

well, I've already stated that I'm [i]on[/i] the cross country team and I posted that my fastest mile so far is a 4:54....so...

Looking at past track results, there are a bunch of people who can run sub 4:20 in the [i]1500[/i].  If you can run 4:20 in the [i]mile[/i] that should definitely put you in the top 7.

To make cross country you have to be able to break 18 in the 5K after the first week of practice and 30 in the 8K after the second week of practice though.

If you're interested in MIT cross country and/or track and field you should [url=http://web.mit.edu/hwtaylor/mtfxc/recruits.html]recruit yourself[/url] ;)

[quote]Thats wierd, you should be able to do much better at 1 and 2 mile (unless those are your splits)
a 16:05 is pretty fast, my pr this season is only 19:05... (but I made the local paper for it because my coach liked my negative splits so much)
and I don't really do track (ran like 3 meets) so out of those my mile time is a sluggish 5:02...[/quote]

I'd guess the 5K time is from this seasons while the 1/2 mi. times are from last season, of course.  He must have put in a good summer of training :)

Hey if I get accepted into MIT I could probably make cross country!
Does MIT give any scholarships (I still have time, i'm a sophomore...)

I dunno...try mitadmissions.org! I was just on there.

The first thing you should know about Ivy and most elite schools is that they only give out need based scholarships.  They calculate how much money they think you and your family can handle to pay and take out with loans and they give you the rest in a scholarship.  But no matter what kind of accomplishments you have you won't get any additional scholarships from the school.  You can always get outside scholarships though.

Okay that's what I had heard but I wasn't sure if it was all ivy's...


calculus
derivative
function
Putnam
geometry
rectangle
3D geometry

does anyone have interesting problems on maxima and minima?

Every calculus textbook has a section on Applications of Differentiation!

1. Find the absolute maximum value of the function: $ g(t) \equal{} \frac{1}{1 \plus{} |t|} \plus{} \frac{1}{1 \plus{} |t\minus{}2|}$

2. Find the local maximum and minimum values for the function: $ f(x) \equal{} e^{\minus{}\frac{1}{x\plus{}1}}$

Here's one I just assigned to my Putnam problems class. It's Putnam 1996 A1:

Find the least number $ A$ such that for any two squares of combined
area 1, a rectangle of area $ A$ exists such that the two squares can
be packed in the rectangle (without interior overlap). You may assume
that the sides of the squares are parallel to the sides of the
rectangle.

The Putman question is very nice.

For 2, it seems that the function approaches $ \infty$ as $ x$ approaches $ \minus{} 1$ from the left and that it approaches $ 0$ as $ x$ approaches $ \minus{} 1$ from the right.

For 1, sketching the graph, considering $ x < 0, x \equal{} 0, 0 < x < 2, x \equal{} 2$ and $ x > 2$, we see that there is a global maximum of $ \frac {4}{3}$ when $ x \equal{} 0$.

3. Prove that a solid right circular cone of given total surface area has its greatest volume when the slant height is three times the radius of the base.


Solve equation
\[ \sqrt {5{x^2} \plus{} 14x \plus{} 9}  \minus{} \sqrt {{x^2} \minus{} x \minus{} 20}  \equal{} 5\sqrt {x \plus{} 1}\]

[quote="litbon"]Solve equation
\[ \sqrt {5{x^2} + 14x + 9} - \sqrt {{x^2} - x - 20} = 5\sqrt {x + 1}\]
[/quote]

The equation is $ \sqrt{(x+1)(5x+9)}-\sqrt{(x+4)(x-5)}=5\sqrt{x+1}$ and we need $ x\geq 5$

We can write the equation as $ \sqrt{x+1}(\sqrt{5x+9}-5)=\sqrt{(x+4)(x-5)}$

Squaring gives $ 2x^2+20x+27=5(x+1)\sqrt{5x+9}$

Squaring again gives $ 4x^4-45x^3+33x^2+505x+504=0$

Which may be written $ (4x^2-25x-56)(x^2-5x-9)=0$

And so four real roots for this equation : $ -\frac 74,\frac{5-\sqrt{61}}2,\frac{5+\sqrt{61}}2,8$

From these, we can only keep the two roots above $ 5$ and it is easy to check back in the original eqsuation that these two roots fit the equation.

Hence the result : $ \boxed{x\in\{\frac{5+\sqrt{61}}2,8\}}$


algebra
polynomial
linear algebra
linear algebra unsolved

I tried to solve this one. But I couldn't get a proof.

[b]Problem: [/b] Find the condition under which $AB=BA$

where matrices $A, B$ have an order ($n\times n$).

Maybe you should state your problem more specifically.
If $AB=BA$ for all $B$, then $A$ must be $\lambda I_{n}$.
If you fix $A$ and want to find all $B$, then I have no idea, even though $det(A)\neq0$.

Thank you for comment again 'cyqwhu'  :D

This problem is about fixed $A$ and find all $B$.

My brother answered this question to me.

I found some conditions for $n=2$ and I wondered whether

there's some generallizations. This is a simple question

and terribly difficult to solve. !

Do you mean the problem is:
find all $B$,such that for every $A$ we have:
$AB=BA$?

In other word,  problem is

Find all pairs $(A,B)$ which satisfy $AB=BA$.

This is the problem.

but what kind of conditions are you looking for ?

i.e. $A$ or $B$ must be represented as polynomial form of other one.

$A=a_{n}B^{n}+a_{n-1}B^{n-1}+\cdots+a_{2}B^{2}+a_{1}B+a_{0}$.


a old man has 3 grandchildren, anna, bob and cathy

the 3 grandchildren live in an area with 20 cities, and each grandchild lives in a different city.

the old man wants to see his 3 granchildren.
He only knows which city anna and bob is in, but he forgets which city cathy lives in.

original: what is the probability that the old man will visit all 3 of his granchchildren after visiting each city exactly once?

edited: what is the probability that the old man must goto all of the cities in order to visit all 3 of his granchchildren?

*sorry, i worded the problem incorrectly.

this problem hit me this morning when i was waking up, so i have no idea what the aswer would be. if you know the answer, then please provide an explanation as well. thanx in advance

Um...if he's visited all the cities, he's already seen his grandchildren.

If he already knows where Ann and Bob live, then he only needs to find Cathy, so $ \dfrac{1}{18}$ probability the next city he visits will be right.

question has been reworded since i actually meant something else, it wasn't suppose to be that easy,
sorry everyone

Okay, so he already knows where Ann and Bob are, so they are discounted.

[hide]He must now visit 18 cities such that he visits Cathy last.  There is a $ \dfrac{18\minus{}n}{19\minus{}n}$ chance that Cathy will not be in the $ n^{th}$ city.  This leads to $ \frac{1}{18}$ after multiplication.[/hide]

Or
[hide]There are $ 18!$ ways to organize the man's trip among the cities.  There are $ 17!$ trips that end with Cathy's city.  Thus there is a $ \frac{17!}{18!}$ or $ \frac{1}{18}$ probability that he must visit all the cities.[/hide]


Hello all,

I am having a problem inserting an image in a specific location. It is automatically placed at the end of the paragraph (on a new page) instead of where I typed the \includegraphics command. This is the code:

[code]
This curve shows the amount of power generated at a specific wind speed. The one used in this project is shown below.
\begin{figure}
  \includegraphics[scale=0.75]{C:/Programme/MATLAB704/work/untitled.png}
  \caption{Machine Curve}
  \label{fig:machinecurve}
\end{figure}
The resulting power is equal to:
\begin{displaymath}
  \displaystyle\int^\infty_{-\infty} f(v) \Phi(v) dv %-inf to inf ??
\end{displaymath}
where $f(v)$ is the computed Weibull fit pdf, and $\Phi(v)$ is the machine curve.
[/code]

... and the result
[img]http://www.hatahet.net/latex.png[/img]

I want the image to appear directly after the statement "...is shown below."

Thanks in advance

It's not the \includegraphics command that's determining where the picture goes, it's the \begin{figure} command.  This creates what is called a "float", and LaTeX will try to place floats wherever it thinks is best, which is not always where you think is best.

In particular, LaTeX will never place a float in the middle of a paragraph, which is what your code is trying to ask it to do.  In fact, LaTeX's default behavior is to try to place floats on the top of the page.

You can try to force a figure to appear exactly where you want it by placing a [h] immediately after the \begin{figure} command:

\begin{figure}[h]
\includegraphcs{.....

(The [h] stands for "here".)

DPatrick,

It's working pretty well, but what should I do if the image is near the end of the page?
[img]http://www.hatahet.net/untitled.png[/img]

...and the code

[code]Once the file is selected, three events occur: the first date in the file is automatically loaded into the \emph{Date} text field; the range of the dates contained in the file is displayed in the prompt; and four pushbuttons appear -- \emph{Wind Rose}, \emph{Speed vs. Time}, \emph{Show CDF}, and \emph{Generate Table}.
\begin{figure}[h]
  \begin{center}
    \includegraphics[height=3.25in]{afterload.png}
  \end{center}
  \caption{Post-file Loading}
  \label{fig:afterload}
\end{figure}
[/code]

Obviously it won't fit at the end of the page, so it must be placed at the beginning of the following page. I tried to insert a "\newpage"... it works, but it messes up all the remaining figures.

Hi,
If you're really intent on showing LaTeX who's boss (and I shamefully admit I have resorted to this when faced with corralling several floats in quick succession), use the option parameter !h instead of h: !h meaning nowhere but here, no matter how crappy it makes my document look.
SSN.

[!h] fixed some issues, but I still have to insert a \newpage manually. Is there a way to make LaTeX automatically insert a page break if the figure doesn't fit on the current page (and have the proceeding text placed after the figure properly)?

Thanks


vector
trigonometry
trig identities
Law of Cosines

2x^2-2y^2+4z^2+6xz+2yz+3yx=0 represent a pair of planes then the angle between pair of plain is

[hide="Solution"]Observe that (x,y)=0 means that z=0, and similarly for x and y. Therefore, the equation must be equivalent to the product of two linear combinations of x, y and z equal to 0.

We can use the template (2x+ay+bz)(x+cy+dz)=0, and solve to find that the equation can be written as (2x-y+2z)(x+2y+2z)=0.

The angle between the two planes is the angle between their normal vectors.

The normal vectors are (2,-1,2) and (1,2,2). Both of these have length 3, by the distance formula.

Their difference is (1,-3,0), which has length sqrt(10)

Now, we can just apply the law of cosines: 18-18c=10, so c=4/9, and the angle is arccos(4/9)[/hide]

Probably would've been somewhat more appropriate in Intermediate Topics.


If there's anyone also working through this (or would like to, as in my limited opinion, it's a pretty cool problem), my answers for parts e and f follow:

Letting counterclockwise be positive, the angular momenta for the smaller and larger cylinders are $\frac{q \mu_0 n a^2 I}{2}$ and $-\frac{q \mu_0 n b^2 I}{2}$. The change in angular momentum for the fields is $q \mu_0 I n \frac{b^2 - a^2}{2}$.

For part E, I think you should get two separate answers for the inner and outer cylinders. I worked through the problem a while ago and dont remember my answers, but I believe the main reason they asked for the flux around the inner and outer cylinders was to see that the student recognizes that there is no magnetic field outside the solenoid and thus you dont include the area outside the solenoid in the magnetic flux of the outer cylinder. Conversely, the area of the solenoid outside the inner cylinder is not included in the magnetic flux calculation for the inner cylinder. Part F, since an addition in angular momentum in one section caused a decrease in another section of the system, the answer would just be the sum of the two expressions obtained in part E. You have to be careful though, since one cylinder spins in the opposite direction as the other cylinder and thus has a negative angular momentum with respect to that cylinder. Sorry if you already recognized this, but I'm too lazy to work through the problem right now and I cant see if your answers are correct or not.

Besides, Part B1 on that exam was a much cooler problem.

sorry, I just gave the momentum for the smaller cylinder ... I edited my post and indeed, the total change in angular momentum is $0$ - yay!


inequalities
linear algebra
matrix
linear algebra theorems

Does anyone know of good uses for the Rational Canonical Form?  I know the Jordan Form, Schur Form, Singular Value Decomposition, etc. all have great uses for solving things like systems of differential equations or have applications in analysis.  Anything that hands you the eigenvalues of a matrix is going to have many applications.  The RCF doesn't seem to give that much away though.  I may be wrong as I have just learned about it.  Any insight into things you have used it for or known uses would be great!  Thanks.

The rational canonical form does not leave the field we started in. It's particularly useful precisely when we [i]don't[/i] have eigenvalues. The applications are to algebra rather than analysis; analysis is done over $\mathbb{R}$ or $\mathbb{C}$, where we always have complex eigenvalues.


Hey, since you guys are all so much better than me at math...

Could you please teach me about Diophantine equations?

I don't want any links to useful websites because I don't really learn anything from those.

You guys provide much simpler explanations that can help me understand it all. :lol:

So can you please help?

Thanks!

Problem: 3x + 5y = 90. What is the number of ordered pairs (x,y) that could satisfy this equation?

IDK!!!!!!!!!!!!!!!!!!!!!!!!!

I assume you mean [i]positive, integer solutions.[/i] Otherwise, there's an infinite amount. :P

The first thing you want to do is choose a variable. As you will see, it's easier to choose the one with the largest coefficient, so in the case of 3x + 5y = 90, we'll pick the $ y$ variable.

The next step is to write $ y$ in terms of $ x$. This gives us $ y \equal{} \frac{90 \minus{} 3x}{5}$.

Now we just keep choosing values of $ x$ that will give us positive, integer values of $ y$ until we run out. The first is $ x \equal{} 25$, which gives us $ y \equal{} 3$. Next is $ x \equal{} 20$, where $ y \equal{} 6$. And so on. We keep going until we get $ x \equal{} 0$, $ y \equal{} 16$. But wait! We need positive integer solutions, so we don't include that, so we stop at (5, 15).

I thought it would be even easier if we simplified the fraction first to
$ 18\minus{}3/5x\equal{}y$.  Then we know we can only have x be a multiple of 5(so y is an integer value) so our only options for x is 5,10,15,20, and 25.  So our solutions are (25,3), (20, 6), (15, 9), (10,12), and (5,15).  I just learned Diophantine equations too from a link provided earlier, and it's not that bad I guess.

[quote]Now we just keep choosing values of x that will give us positive, integer values of y until we run [/quote]out.

Isn't that basically just guess-and-check?[/quote]

No. You start with the largest possible value of $ x$ that will give us a $ y$ bigger than 0, which is 29. But we need $ y$ to be an integer, so $ x \equal{} 29$ won't work. However, $ x \equal{} 25$ does, since $ 90 \minus{} 3(25)$ is divisible by 5. From there, you decrease the $ x$ value by $ 5$, and then stop when we run out.

If the coefficient of the variables are 1 than there is an easier way.
For example:
What is the possible positive integer solutions
$ x\plus{}y\plus{}z\equal{}100$
You can just apply balls and urns from this point. Think of it as you have a hundred balls and 3 urns to put them in. But since each x,y,and z have to be greater than 0 lets distibute 1 ball to each of them. So we have 97 balls and three urns. From here we just apply the formula:
$ \binom{97}{3\minus{}1}$

Also to get a few positive solutions to get you started you can take it all modulo the smaller coefficient.


geometry
trapezoid
analytic geometry
graphing lines
slope
perpendicular bisector

you can draw 2 right triangles, with hypotenuses r and legs of 4, 3, and other legs of x+1, x

the triangles start at the center, have a point on the endpoint of one of the chords, and a point on the midpoint.

i think there are 2 possiblies, one where the diameter is between the chords and one where it is under.

Wow ... that's a lot simpler than what I did.  It works out well.

     -- Pawel

Actually, there are three possibilities: the two you named, and one where the eight-inch chord is a diameter itself.

but if the 8 is the diameter, the chord 1 cm above does not have length 6, it has length $\sqrt 15$

[hide]Draw the radii to the intersections of the chords with the circle. Draw the diameter perpendicular to the two chords. This diameter is also the perpendicular bisector of the chords. Now you have two right triangles. Let $x$ be the distance from the length 8 chord to the center. Both chords are on the same side of the circle; otherwise, you get negative distances.
$r^2=x^2+16\\r^2=\left(x+1\right)^2+9\\x=3$
Then use chord-chord on the perpendicular diameter with the length 8 chord. $\left(r+3\right)\left(r-3\right)=4^2\to r=5$[/hide]

If the length 8 chord is the diameter, you get 1-3-4 right triangles. :rotfl:

Edit: Was this my 100th post? Yay, Poincare conjecture!

Ooh, yes, you're right. I didn't think about that.

[quote="verneer"]Here is the problem:

Two parallel chords of length 6 inches and 8 inches are 1 inch apart in a circle.  Find the diameter of the circle.

I solved it by placing the trapezoid on the Cartesian Plane and then working with slopes, points, etc.  Is there a simpler geometric solution to this problem?

     -- Pawel[/quote]

Can you post your oringinal solution?? I want to see a analytic solution since i am not very familiar with it.


geometry
circumcircle
angle bisector
geometry proposed

Let $ \triangle ABC$ be a triangle with incircle $ \omega(I,r)$and circumcircle $ \zeta(O,R)$.Let $ l_{a}$ be the angle bisector of $ \angle BAC$.Denote $ P\equal{}l_{a}\cap\zeta$.Let $ D$ be the point of tangency $ \omega$ with $ [BC]$.Denote $ Q\equal{}PD\cap\zeta$.Show that $ PI\equal{}QI$ if $ PD\equal{}r$.

From angles $ \angle DQB \equal{}\angle PQB \equal{}\angle PAB \equal{}\angle PAC \equal{}\angle PBC \equal{}\angle PBD$ So circumcircle of $ \bigtriangleup QDB$ is tangent to $ PB$. From tangent teorem and $ PB \equal{} PI$ (it is well-known) we get $ PD\cdot PQ \equal{} PB^{2}\equal{} PI^{2}\Rightarrow$ circumcircle of $ \bigtriangleup DIQ$ ıs tangent to $ PI\Rightarrow\angle IQP \equal{}\angle IQD \equal{}\angle DIP \equal{}\angle IPD \equal{}\angle IPQ\Rightarrow IP \equal{} IQ$

Dear,cefer,Did you take participant in this olympiad,i mean not exactly in 2002's,but in SRMC at all.Thank you for your nice solution.

I participated in 2005-2006, but i dont know my teachers sent our results or not. I think in 2006 they were late for sendıng answers.

[b]Very nice problem and the Cefer's proof ! Here is another proof.[/b]

[quote="Erken"][color=darkred]Let $ ABC$ be a triangle with the incircle $ \omega(I,r)$ and the  circumcircle $ \Omega (O,R)$. Let $ l_{a}$ be the angle bisector of the angle $ \widehat{BAC}$.  Denote $ P \equal{} l_{a}\cap\Omega$. Let $ D$ be the point of tangency of the circle $ \omega$ with  the side $ [BC]$. Denote the second intersection $ Q\in PD\cap\Omega$. Show that $ PD \equal{} r\implies IQ \equal{} IP$.[/color][/quote]
[color=darkblue][b]Proof.[/b] Denote the second intersection $ A'\in AO\cap\Omega$. Since $ \widehat{DIP}\equiv\widehat{IAA'}$ and the power $ p_{\Omega}(I)$ of the point $ I$ w.r.t. the circle $ \Omega$ is $ \minus{}p_{\Omega}(I) \equal{} AI\cdot IP \equal{} 2Rr \equal{} ID\cdot AA'$, i.e. $ \frac{ID}{IP}\equal{}\frac{AI}{AA'}$ results $ \triangle IDP\sim\triangle AIA'$. Thus $ \widehat{IPD}\equiv\widehat{AA'I}$ what means that the point $ \boxed{\ Q\in PD\cap IA'\ }$. Therefore, $ \widehat{DQI}\equiv\widehat{IAA'}\equiv\widehat{DIP}$. In conclusion, $ PD \equal{} r$ $ \Longleftrightarrow$ $ PD \equal{} DI$ $ \Longleftrightarrow$ $ \widehat{DPI}\equiv\widehat{DIP}$ $ \Longleftrightarrow$ $ \widehat{DPI}\equiv\widehat{DQI}$ $ \Longleftrightarrow$ $ IP \equal{} IQ$.

[b]Remark.[/b] $ \boxed{\ PD \equal{} r\Longleftrightarrow IP \equal{} IQ\Longleftrightarrow IO\perp IA\Longleftrightarrow b\plus{}c \equal{} 2a\ }$.
[/color]

[quote="Erken"]Where must your teacher sent your results?I suppose that in your country,as in mine,APMO and SRMC come together,APMO results are sent to Korea,but i don't know about SRMC.. :maybe:[/quote]

I think they send SRMC results to Turkey


A circle is inscribed in a triangle with sides 8, 15, and 17. The radius of the circle is :

a. 6
b. 2
c. 3
d. 5

[quote="archimedes1"][hide]We have two ways of finding the area of a triangle: $ A=\sqrt{s(s-a)(s-b)(s-c)}$ (Heron's semiperimeter formula), and $ A=rs$.  Setting these equal and solving for $ r$, we find
$ r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$.[/hide][/quote]

[hide]
I'll finish this off.

since $ s=\frac{8+15+17}{2}=20$, and $ r=\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}$:

$ r=\frac{\sqrt{20\cdot 12 \cdot 5 \cdot 3}}{20}$
$ r=\frac{60}{20}$
$ \boxed{r=3}$

$ \boxed{C}$[/hide]

[hide]Because $ 8^{2}+15^{2}=17^{2}$, it is a right triangle. Then you can find the area is 60.

The formula for the inradius is $ \frac{A}{s}$ (area over semiperimeter).
$ \frac{A}{s}=\frac{60}{20}=\boxed{3}$[/hide]

[quote="archimedes1"][hide]We have two ways of finding the area of a triangle: $ A=\sqrt{s(s-a)(s-b)(s-c)}$ (Heron's semiperimeter formula), and $ A=rs$.  Setting these equal and solving for $ r$, we find
$ r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$.[/hide][/quote]
Shouldn't it be $ r=\frac{\sqrt{(s-a)(s-b)(s-c)}}{s}$?

Yes, I think that was a mistake by archimedes. I fixed that in my solution.

*clears throat*

$ \sqrt{\frac{(s-a)(s-b)(s-c)}{s}}=\sqrt{\frac{s(s-a)(s-b)(s-c)}{s^{2}}}= \frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}$

buzzer11's and my expressions are equivalent.  I think Quevvy forgot an $ s$ in the numerator.


Sorry if this has been asked already.  I'm new to this forum and a cursory search didn't turn up anything related.

My school requires that Appendix appear in the TOC without a dotted leader.  The actual appendices then follow with dotted leaders just like chapters.  I accomplished this by modifying a document class so that \appendix is defined as follows:

[b]EDIT: I simplified this part a lot by using \addtocontents instead of \addcontentsline[/b]

[code]\def\appendix{
  \addtocontents{toc}{\noindent APPENDIX \hfill}
  \setcounter{chapter}{0}
  \setcounter{section}{0}
  \def\@chapapp{APPENDIX}
  \def\thechapter{\Alph{chapter}}
      }[/code]

It works and prints "APPENDIX" in the TOC as desired, but for some reason the first appendix appears before this heading.  I can't figure out why the .toc file is being generated with the following:

[b]EDIT: It still accomplishes what I want, but now it's much simpler.  Somehow the compiler is still adding the entry for Appendix A to the TOC before "APPENDIX"[/b]

[code]\contentsline {chapter}{\numberline {\hfill A.\nobreakspace {}\nobreakspace {}}Correspondence of 2-level Operators to Pauli Matrices}{45}
\contentsline {chapter}{\numberline {\hfill A.\nobreakspace {}\nobreakspace {}}Correspondence of 2-level Operators to Pauli Matrices}{45}
\noindent APPENDIX \hfill
\contentsline {chapter}{\numberline {\hfill B.\nobreakspace {}\nobreakspace {}}Computer Aided Design of Apparatus}{49}
\contentsline {chapter}{\numberline {\hfill C.\nobreakspace {}\nobreakspace {}}Gravitational Tidal Force on a Micro-dimple}{52}
\contentsline {section}{\numberline {\hfill C.1.\nobreakspace {}\nobreakspace {}}Magnitude of the Tidal Force}{52}
\contentsline {chapter}{\numberline {\hfill D.\nobreakspace {}\nobreakspace {}}Acronyms}{54}
\contentsline {spchapter}{BIBLIOGRAPHY}{55}[/code]

Notice that the "APPENDIX" heading follows "A".  Any help would be much appreciated.

I found a way to get it to render correctly, but it seems a bit like a hack.

In my main document I previously had made the following calls:

[code]
\include{chapter1} %import chapter1.tex
\include{chapter2} %import chapter2.tex
\include{chapter3} %import chapter3.tex
\include{chapter4} %import chapter4.tex
\appendix
\include{AppendixA} %import appendixA.tex
\include{AppendixB} %import appendixB.tex
\include{AppendixC} %import appendixC.tex
\include{Acronyms}
\bibliography{GetCitation}
\end{document}
[/code]

This had been generating a faulty TOC with "APPENDIX" being printed on the line after the entry for Appendix A.
I have not dug into the details for why this might be so, but apparently there is a significant difference between the "include" and "input" commands.

I changed the first "include" statement after "appendix" into an "input" and everything works properly.
[code]
\include{chapter1} %import chapter1.tex
\include{chapter2} %import chapter2.tex
\include{chapter3} %import chapter3.tex
\include{chapter4} %import chapter4.tex
\appendix

\input{AppendixA} %import appendixA.tex
\include{AppendixB} %import appendixB.tex
\include{AppendixC} %import appendixC.tex
\include{Acronyms}
\bibliography{GetCitation}
\end{document}[/code]

Out of curiosity, does anyone know why this is the case?


LaTeX
algebra
polynomial
geometry
3D geometry
quadratics
SFFT

\[ \text{Basic Factorizations}  \]

\begin{eqnarray*} x^2 - y^2 &=& (x-y)(x+y)\\ x^3 - y^3 &=& (x-y)(x^2+xy+y^2)\\ x^3 + y^3 &=& (x+y)(x^2-xy+y^2)\\ x^k-y^k &=& (x-y)\left(x^{k-1}+x^{k-2}y+\cdots+xy^{k-2}+y^{k-1}\right)\\ x^{2k+1} + y^{2k+1} &=& (x+y)\left(x^{2k}-x^{2k-1}y+\cdots-xy^{2k-1}+y^{2k}\right)\\ xy + kx + jy + kj &=& (x+j)(y+k)\\ \end{eqnarray*}

The last one is known around AoPS as Simon's favorite factoring trick.  I put a lot of variables in there for the generalization, but that's probably a little hard to understand.  Instead, just remember that $xy+x+y+1=(x+1)(y+1)$.

\[ \text{Problems}  \]

The below problems exhibit other factorization techniques that aren't necessarily listed above.  I hid all of the solutions so that you can try the problems before looking at the solutions.

[b]Problem 1[/b] Factor $n^4+4$.
[hide="Solution"] We use the tactic of cleverly adding zero:

\begin{eqnarray*} n^4 + 4 &=& (n^4+4kn^2+4)-4n^2\\ &=& (n^2+2)^2-(2n)^2\\ &=& (n^2+2n+2)(n^2-2n+2) \end{eqnarray*}[/hide]

[b]Problem 2[/b] Factor \[ x^{10}+2x^9+3x^8+4x^7+5x^6+6x^5+5x^4+4x^3+3x^2+2x+1.  \]
[hide="Solution"]
Let's start off with some smaller things and see if we can find a generalization.  We notice that:

\begin{eqnarray*} x^2+2x+1 &=& (x+1)^2\\ x^4+2x^3+3x^2+2x^2+1 &=& (x^2+x+1)^2 \end{eqnarray*}

Hmm, that's interesting.  So we guess that our expression factorizes as \[ (x^5+x^4+x^3+x^2+x+1)^2.  \]

Indeed, we can expand this and see that it works.  Additionally, we see an immediate generalization: \[ x^{2n}+2x^{2n-1}+\cdots+nx^{n+1}+(n+1)x^n+nx^{n-1}+\cdots+x+1=(x^n+x^{n-1}+\cdots+1)^2.  \]

Roots of unity are also related to this factorization...can you see what I am talking about :)[/hide]

[b]Problem 3[/b] Factor $z^7+z^2+1$.
[hide="Solution"].  Consider what would happen if $z^3=1\Leftrightarrow z^3-1=0=(z-1)(z^2+z+1)$.  Then since $z$ is obvioulsy not equal to 1, we must have $z^2+z+1$.  But if $z^3=1$ then $z^7=z$ and the expression we are trying to factor becomes $z^2+z+1$.  Therefore, we see that $z$ satisfies $z^3=1$ (i.e. $z$ can be one of the non real roots of unity).  So we know that $z^2+z+1$ divides $z^7+z^2+1$.

The rest is just long division giving us

\[ (z^2+z+1)(z^5-z^4+z^2-z+1).  \][/hide]

[b]Problem 4[/b] If $m$ and $n$ are integers such that $m^2 + 3m^2n^2 = 30n^2 + 517$. Find $3m^2n^2$.
[hide="Solution"]
You know I had to throw one in with Simon's favorite factoring trick :D

So make it

\begin{eqnarray*} 3m^2n^2+m^2-30n^2=517\\ \Leftrightarrow (m^2-10)(3n^2+1)=517-10\\ \Leftrightarrow (m^2-10)(3n^2+1)=3\cdot 13^2 \end{eqnarray*}

We see that we must divide 507 up into two factors in such a way that one is 10 less than a perfect square.  This leads us to $m^2-10=39$ and $3n^2+1=13$ from which we get $m=7$ and $n=2$.  Thus $3m^2n^2=588.$[/hide]

\[ \text{Some Additional Problems}  \]

1.  Factor $a^3+b^3+c^3-3abc$.

2. Factor $(a+b+c)^3-(a^3+b^3+c^3)$.

3. Factor $z^{10}+z^5+1$.

4. Factor $(x+y)^7-(x^7+y^7)$.

5.  Find the smallest positive integer $n$ for which the expansion of $(xy - 3x +7y - 21)^n$, after like terms have been collected, has at least 1996 terms.

6. Find the value of (no calculator) \[ \frac{(10^4 + 324)(22^4 + 324)(34^4 + 324)(46^4 + 324)(58^4 + 324)}{(4^4 + 324)(16^4 + 324)(28^4 + 324)(40^4 + 324)(52^4 + 324)}.  \]

Can someone post some solutions?  These are good problems, but some of them stumped me.  Help!

I brute forced number 1...
[hide]
I multiplied out $(a+b+c)^2$ and then subtracted out $(a^2+b^2+c^2)$.  My result was $3(ba^2+ca^2+b^2a+c^2a+2abc+bc^2+b^2c)$.  I'm sure this can be simplified further, however...
[/hide]

[hide="Number 6"]
Joe said we can do this using the Sophie Germain identity, $a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)$.  But I am reluctant to use that on such a long problem.  Is there an easier way?
[/hide]

[hide="1"]The first one is $(a+b+c)[(a+b+c)^2-3(ab+bc+ac)]$
[/hide]
[hide="2"]$3(a+b)(b+c)(a+c)$[/hide]
[hide="3"]
Since it's complex numbers $z^{10}+z^5+1=0$ and $z^{15}+z^{10}+z^5=0$, so our answer is $z^{15}-1$.[/hide]

[hide="4"]Use Binomial Theorem..don't feel like doing it[/hide]

[hide="5"]Simplify it to $((x+7)(y+3)-42)^n$ and work with that, since for every $n$ power on binomials, there are $n+1$ terms, yet there are four extra ones for each power in this case. We have $\frac{1996}{4}=499$, so we need it to the $498$ power which is $n$, did i do that right?[/hide]

[hide="6"]
Using Sophie German, we know that $x^4+324=((x+3)^2+9)((x-3)^2+9)$, so therefore after rewriting all the terms in that form:
$\frac{(7^2 + 3^2)(13^2 + 3^2)(19^2 + 3^2)(25^2 + 3^2)(31^2 + 3^2)(37^3 + 3^3)(43^2 + 3^2)(49^2 + 3^2)(55^2 + 3^2)(61^2 + 3^2)}{(1^2 + 3^2)(7^2 + 3^2)(13^2 + 3^2)(19^2 + 3^2)(25^2 + 3^2)(31^2 + 3^2)(37^2 + 3^2)(43^2 + 3^2)(49^2 + 3^2)(55^2 + 3^2)}$ we see that a lot of terms cancel out and we're left with $\frac{61^2+9}{1^2+9}=\boxed{373}$
[/hide]

Those are all of them although i didn't feel like working out number 4

Shobhit,

In the solution formula for number 6, did you mean
$x^4+324=((x+3)^2+9)((x-3)^2+9)$?

I was working through it and I see how your final answer works, but I couldn't figure out how you derived the general formula; I think you may have accidentally swapped the addition and subtraction signs at the end of the formula.

[quote="drummerdude413"]Shobhit,

In the solution formula for number 6, did you mean
x^4+324=((x+3)^2+9)((x-3)^2+9)?

I was working through it and I see how your final answer works, but I couldn't figure out how you derived the general formula; I think you may have accidentally swapped the addition and subtraction signs at the end of the formula.[/quote]

yeah, sorry, latex slip up, and I will post the derivation of it in a little while (homework :( ), but essentially, you use completing the square and stuff.

EDIT:

Okay, here's the derivation:

first of all, notice that $324=18^2$<----yeah, i know, takes a while to figure that out

from here, we have $x^4+18^2$, now let's do something with completing the square, it can be $(x^2+18)^2-36x^2=(x^2+18)^2-(6x)^2$, now use difference of squares to get: $(x^2+6x+18)(x^2-6x+18)=(x^2-6x+9+9)(x^2+6x+9+9)$, now complete the square again, and we get $((x+3)^2+9)((x-3)^2+9)$...and there you go.

Thanks for that extra explanation, it was what I needed!

[quote="Iversonfan2005"][quote="chess64"][quote="Iversonfan2005"]first of all, notice that $324=18^2$<----yeah, i know, takes a while to figure that out[/quote]

... wait i don't get it are you being sarcastic or something? :?[/quote]

no, did you notice that at first glance? ;)

or maybe i'm just a little slow :|[/quote]

umm yeah sorry i guess i assumed everyone had their squares memorized to at least 30^2, apparently that's not the case ;)

[quote="Iversonfan2005"][quote="chess64"]
umm yeah sorry i guess i assumed everyone had their squares memorized to at least 30^2, apparently that's not the case ;)[/quote]

heh, yeah right :P  :roll: , the day i memorize my squares is the day I make IMO<------in that case, maybe I should start right now  :rotfl:[/quote]

it's not hard to memorize squares... it's just like 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, ...

[quote="agolsme"]I am sort of confused about the example problem #3. Why do we start with the assumption that $z^3=1$?[/quote]

cuz the maker of the problem told me it's complex numbers and stuffs

but, how does that mean z^3 = 1? I mean, how did you know that z was that particular complex number?

Instead of polynomial expressions, let's look at factoring positive integers instead.  How do you factor an integer?  Unless you can immediately see some 'trick' (i.e. divisibility rule) you are left with guess and check.  The same goes for polynomial expressions.  Unless you can see some 'trick' (like difference of squares, sum of cubes, etc) you are left guessing and checking.  Even factoring quadratics you usually have to guess and check (although you probably do this mentally).  That's basically what we're doing here.  We don't see any immediate tricks so we start guessing.  We guess that $z^2+z+1$ or $z^3-1$ for $z\neq 1$ is a factor.  This might seem a little random at first but it's something you should look for.  And it's really not that hard to check since you can use the fact that $z^3=1$ to simplify high powers.  If you can get it down to $z^2+z+1$ you know that $z^2+z+1$ is a factor.

This works for other $n$-th roots of unity.  For example, the 4th roots of unity you would have $z^4=1$ and thus could simplify from there.

Does that make it seem a little more natural?  We only assumed that $z^3=1$ in that problem in the same sense that we assume 7 divides 91 before we test it out.


Reactions that involve rearrangements are not only interesting from a mechanistic point of view, but can also be very usefull in synthesis. Let's take a look at some important examples.

1) In each of the following, find the major product that is expected to be obtained and explain your answer with a mechanism:

(a) 2-Aminocyclohexanol + nitrous acid ----->

(b) 1-Phenyl-1,2-ethanediol + sulphuric acid ----->

(c) 2-Diazocyclohexanone + methanol/light ----->

(d) p-Methylbenzamide + NaHO/bromine ----->

(e) 3-Bromobenzoic acid + hydrazoic acid/heat ----->

(f) Cyclohexanone + hydroxylamine -----> A , then A + sulphuric acid ----->

(g) Acetophenone + hydrazoic acid/sulphuric acid ----->

(h) Acetophenone + CF3CO3H ----->

(i) Phenyl hydroxylamine ($ C_6H_5NHOH$) + H3O+ ----->

Yawn I'm going to go to sleep but I think I'll just have time enough to write one answer
i)
C6H5NHOH -----> C6H4NH2OH
                 H+
i.e. OH in the posn para to NH2.
Its called the para amino phenol rearrangement.

It is more commonly known as the Bamberger Rearrangement.

d. You get para toluidine
    Its called the Hoffmann Bromamide reaction

f. 1st you get cyclohexaneoxime which later rearranges to form caprolactam
Its called the Beckmann Rearrangement

h. Its Baeyer Villiger oxdn  :wink:

And by the way isn't addn of $ N3H$ called the Schimdt rearrangement ?  :maybe:

[quote="Madness"]d. You get para toluidine Its called the Hoffmann Bromamide reaction[/quote]

That's right, but the reaction's name is Hoffmann Rearrangement.

[quote="Madness"]f. 1st you get cyclohexaneoxime which later rearranges to form caprolactam
Its called the Beckmann Rearrangement[/quote]

That's right. This reaction is used in industry in the synthesis of Nylon 6.

[quote="Madness"]h. Its Baeyer Villiger oxdn[/quote]

That's right, but what is the product obtained?

[quote="Madness"]And by the way isn't addn of  called the Schimdt rearrangement[/quote]

Yes, but the product obtained depends on the substract.


[quote="Beat"]$x+y+z=1$ are positive real numbers. Prove that $\frac{1}{1-xy}+\frac{1}{1-yz}+\frac{1}{1-zx}\le \frac{27}{8}$[/quote]
$\frac{1}{1-xy}+\frac{1}{1-yz}+\frac{1}{1-zx}\le \frac{27}{8}\Leftrightarrow$
$\Leftrightarrow8\cdot\sum_{cyc}(1-xy)(1-xz)\leq27(1-xy)(1-xz)(1-yz)\Leftrightarrow$
$\Leftrightarrow3-11(xy+xz+yz)+19xyz-27x^{2}y^{2}z^{2}\geq0\Leftrightarrow$
$\Leftrightarrow3(x+y+z)^{2}-11(xy+xz+yz)(x+y+z)+18xyz+xyz(1-27xyz)\geq0\Leftrightarrow$
$\Leftrightarrow2\cdot\sum_{cyc}(x^{3}-x^{2}y-x^{2}z+xyz)+\sum_{cyc}(x^{3}-xyz)+xyz(1-27xyz)\geq0.$ :)
The following inequality is stronger:
Let $x,$ $y$ and $z$ are positive numbers such that $x+y+z=1.$ Prove that
$\frac{1}{2-3xy}+\frac{1}{2-3yz}+\frac{1}{2-3zx}\le \frac{9}{5}$

I don`t know why $\sum_{cyc}(x^{3}-x^{2}y-x^{2}z+xyz)+xyz(1-27xyz)\ge 0$
Show me please

The first one is Schur. The second one is because
$1 = (x+y+z)^{3}\ge (3\sqrt[3]{xyz})^{3}= 27xyz$
by AM-GM.

[quote="Beat"]prove me schur  :blush:   :blush:[/quote]
$\sum_{cyc}(x^{3}-x^{2}y-x^{2}z+xyz)\ge 0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(2x^{3}-x^{2}y-x^{2}z)-\sum_{cyc}(x^{2}y+x^{2}z-2xyz)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(x^{3}-x^{2}y-xy^{2}+y^{3})-\sum_{cyc}(x^{2}z-2xyz+y^{2}z)\geq0\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(x-y)^{2}(x+y-z)\geq0.$
Let's assume $x\geq y\geq z.$ Then $x+y-z\geq0,$  $x+z-y\geq0$ and  $(x-z)^{2}\geq(y-z)^{2}.$
Hence, $\sum_{cyc}(x-y)^{2}(x+y-z)\geq$
$\geq(x-y)^{2}(x+y-z)+(y-z)^{2}(x+z-y)+(y-z)^{2}(y+z-x)=$
$=(x-y)^{2}(x+y-z)+2z(y-z)^{2}\geq0.$ :)


calculus
integration
analytic geometry
algebra
polynomial

How many different point in three dimensional space have three positive integeral coordinates whose sum is 100?

I think it's 99 times 98 divided by 2, but i dont know where that comes from.

Suppose you pick x first
100, (y,z)=(0,0)
99 - (y,z)=(1,0) or (0,1)
98 - (0,2) (1,1) (2,0)
so there are in total 1+2+3+...+100 choices, which I think the solution is 100x101/2=5050

RobRoobiks has the right idea, except $ x$ must be strictly positive, so you remove his addition of $ 100$. Also, you remove $ 99$ because when $ x \equal{} 1$, there are only $ 98$ possibilities for $ y$. Therefore, $ \displaystyle\sum_{i\equal{}1}^{98}i \equal{} \frac{98\times99}{2}$.

You can also use the classic balls and urns argument.

We need $ x\plus{}y\plus{}z\equal{}100$, solved in the positive integers. List out a series of objects and dividers to graphically represent a solution. For example, \[ ****|**|***\cdots*\] would describe the solution $ x\equal{}4$, $ y\equal{}2$, $ z\equal{}94$

There are $ 99$ slots in between the $ 100$ $ *$s, and we need to choose two of them to be $ |$s, so $ \binom{99}{2}\equal{}\frac{99\cdot 98}{2}\equal{}\boxed{4851}$.

[quote="DaReaper"]ohhh nice. i like that method. is this just like finding the number of coefficients in the expansion of say, (2x+3y+4z)^3?[/quote]

The "number of coefficients"?

Well, if you want to think in terms of polynomials, we can describe the answer using the idea of generating functions. Basically, we find a polynomial whose coefficients give us our answer. Each variable contributes anywhere from $ 1$ to $ 99$, and there are $ 3$ variables, so our answer is the coefficient of $ x^{100}$ in $ (x\plus{}\cdots\plus{}x^{99})^3$. (Think about why)

[quote="DaReaper"]ohhh nice. i like that method. is this just like finding the number of coefficients in the expansion of say, (2x+3y+4z)^3?[/quote]

Yes, that is the same method. Let $ a$ be the power of $ x$, $ b$ be the power of $ y$, and $ c$ be the power of $ z$. We want to compute the number of distinct nonnegative integral triplets $ (a, b, c)$ such that $ a\plus{}b\plus{}c\equal{}n$, where $ n$ is the power we're taking the trinomial to, and we would apply the above method. However, make sure you account for the possibility that some of $ a$, $ b$, and $ c$ can be zero.


irrational number
number theory unsolved
number theory

Prove that for each natural number $k>1,$ there exists an irrational number $r$ such that for every natural number m, $[r^{m}]-1$ is divisible by $k$.

Let $r>1,r+\frac{1}{r}=a\in N$, then $[r^{m}]-1=r^{m}+r^{-m}-2=a_{m}-2$.
We have $a_{0}=2, a_{1}=a, a_{n+2}=aa_{n+1}-a_{n}$. Therefore $k|(a_{n}-2)\ \forall n$ if $a=2(mod k)$. We can chose a>2, suth that $D=a^{2}-4$ is not perfect square. It give needed r: $r=\frac{a+\sqrt D }{2}$.

sorry but cud u pleeeees b a bit more clr. i dint undrstand the soln
sumbdy pleeees explain this soln to me.
thanks in advance

er er im sorry im reallly sorry
in the question theres a major correction:
Prove that for each natural number $k\geq 1$ there exists an irrational number $r$ such that for every natural number m,$[r^{m}]+1$  is divisible by $k$.

I found r(k), suth that $k|[r^{m}]-1 \ \forall m\in Z$, were $r=\frac{a+\sqrt D }{2}, D=a^{2}-4, a>2 \ a=2(mod k)$, for example $a=2k+2$ give $r=k+1+\sqrt{k(k+2)}$.


in any triangle ABC , h_i are the altitudes
prove

\[ h_a  \plus{} h_b  \plus{} h_c  \le \frac{{\sqrt 3 }}{2}\left( {a \plus{} b \plus{} c} \right)
\]

[quote="apratimdefermat"]in any triangle ABC , h_i are the altitudes
prove
\[ h_a \plus{} h_b \plus{} h_c \le \frac {{\sqrt 3 }}{2}\left( {a \plus{} b \plus{} c} \right)
\]
[/quote] we must prove $ asinC\plus{}bsinA\plus{}csinB\le \frac{{\sqrt 3 }}{2}(a \plus{} b \plus{} c)$
$ sin"x\le 0$ so by Karamata $ ((A,B,C)\succ (\frac{\pi}{3},\frac{\pi}{3},\frac{\pi}{3}))$
$ asinC\plus{}bsinA\plus{}csinB\le \frac{{\sqrt 3 }}{2}a \plus{} \frac{{\sqrt 3 }}{2}b\plus{} \frac{{\sqrt 3 }}{2}c$

can you kindly say what is karamata???actually i havnt seen this term yet...may be it has another name...

hello, see here:
http://www.mathlinks.ro/viewtopic.php?t=231490&search_id=93873285
Sonnhard.


ev danas bilo drzavno iz fizike. Ne znam rezultate. Samo znam da je prvih 5 ucenika (5 takmicara i ide na IPHO) bilo iz Federacije pa ce tih 5 ucenika MORATI RADITI PONOVO jer jedan ucenik mora bit iz Republike Srpske...

SRAMOTA

Samo se pitam do kad ce mjesto stanovanja a ne znanje odredjivati plasman na takmicenju u BIH.

Here we go again. Žao mi je da se ovo ponavlja iz godine u godinu, jer se ista stvar i meni desila 2007, ali sam za svoje dobro odustao od IPhOa na koji sam se plasirao, da se ne bih prepirao sa poštovanim Ignjatovićem i Musemićevom koji su osjećali za prijeko potrebno da umjesto petoplasiranog (mene), ode desetoplasirani sa podobnom adresom...

Tja, baš su me jučerašnji rezultati obradovali (sa porukom, bravo Edina!  :winner_first: ), a sad sam već ljut...

Mi nemamo takvih problema, ali imamo jedan drugi...

Kod nas na IPhO mora ici bar jedan ucenik koji nije iz specijalnog odeljenja, a na IJSO 2!!! Zar drzavu ne predstavljaju najbolji??? Svi su imali priliku da se upisu u specijalna odeljenja...

Meni svejedno, idem na IJSO svakako :)

Hm, i to zvuči poprilično besmisleno. Stvarno bi ljudi koji postavljaju pravila za takmičenja mogli uključiti centar za logiku u malom mozgu... Mada je to kod vas malkice normalnije, jer se pravila postavljaju prije takmičenja, a ne nakon izrade zadataka i objave rezultata kao kod ovih naših fizičara.
Mi svoj problem možemo riješiti, problem je u par ljudi koji prave gluposti 'jer im se može'. I sad se sjećam zvorničke ekipe na našem državnom iz fizike 2007 i njihove podrške mojoj malenkosti u borbi protiv S. Ignjatovića i njegove potrebe da pravi nacionalno podobnu ekipu...

Pa, ekipe za IJSO, IPHO?

Srbija - IJSO

1. Milan Krstajic
2. Teodora Djordjevic
3. Ljubica Vujovic
4. Dusan Sobot ( :) )
5. Adis Dijab
6. Kristina Dzodic

Jos se ne zna ekipa za IPHO, ima jos jedan selekcioni proces.

U prošlom postu zaboravih da ti čestitam plasman na IJSO  :)  naravno, ime koje mi je najzanimljivije na spisku jeste Mladen Krstajić  :lol: (inače, za one koji ne znaju, onaj malo poznatiji Mladen Krstajić je rođeni Zeničanin, pa možda i ovaj vuče korijene od janjićkih Krstajića  :lol: )

Što se teme tiče, news update: kako saznajem, prvih troje sa državnog takmičenja ide na IPhO. S njima ide i sedmoplasirani (iz RS), a dvoje koji su podijelili četvto-peto mjesto na državnom takmičenju, razigravaju za preostalo mjesto u ekipi.

Jao izivinjavam se, nije Mladen, vec Milan  :)

Hvala!

A ko čini tu našu (ne)srećnu šestorku.Da li je balkanijac :)  Alija Bevrnja među njima

Zukić Edina, Kohnić Arman, Bevrnja Alija su zauzeli prva tri mjesta na državnom. Ko je član ekipe iz RSa ne znam, a nakon razigravanja ćemo saznati i petog člana (IPhO timovi imaju pet članova).

Zaista nemate pojma o pravilima formiranja ekipe BiH za IPho.
Pravilo govori da ekipu moraju sacinjavati ucenici iz oba entiteta...
A u ostalom, primjedbe na dr Sinisu Ignjatovica i dr Rajfu Musemovic su vam zaista primitivne!!!

Hvala na ukazivanju na moj primitivizam, uzeto k znanju - razmišljam o samobaniranju. Drago mi je da je sve ovo zasnovano na pravilima, ali ja imam par pitanja vezanih za taj pravilnik:

1. Koliko je taj pravilnik star?
2. Ko je sve upoznat sa tim pravilnikom?

Ovo pitam jer se na državnom takmičenju 2007. desilo sljedeće:

-na početku takmičenja je saopćeno da prvih pet sa tog takmičenja ide na IPhO u Iran.
-izlaze preliminarni rezultati.
-izlaze konačni rezultati
-počinje farsa:

+parafraziram profesoricu Musemić na zatvaranju takmičenja: Na olimpijadu zbog loših rezultata četverto- i petoplasiranih idu samo prvih troje. Profesorica izražava žaljenje što u ekipi nema nikog iz RSa, te nagovještava mogućnost da se nešto 'sredi', jer, Bože moj, takmičari iz RSa nisu na vrijeme obaviješteni o terminu državnog.

+naknadno, na Olimpijadu odlazi i četvrtoplasirani, a da se ekipa popuni, ekipa se čini multinacionalnom.

+niko, apsolutno niko ne govori o tom pravilu po kom na Olimpijadu mora ići kombinirana ekipa.

mikky333 ja stvarno ne razumijem gdje ti vidis logiku u tom pravilniku :/ valjda treba da postoji samo jedno pravilo a to je da najbolji predstavljaju zemlju tj da udju u reprezenataciju.
A 4. mjesto na takmicenju osvojio je Husremovic Faruk Prva Gimnazija Zenica =)

i mene takodjer zanimaju odgovori na pitanja koja je postavio gdin hsiljak. I zaista ne razumijem kako ljudi koji su na tako odgovornim pozicijama nece da vide besmislenost ovog sto se sad desava. Bosna je jedna i svi smo mi bosanci =)

Pa, taj pravilnik je prihvatilo drustvo fizicara BiH.
I ja sam sad bio ucesnik - takmicar na ovogodisnjem
Drzavnom takmicenju. Plasirao sam se negdje u sredinu, mada imam kritika.
Ucenici iz RS ne rade zadatke sa integralima i diferencnim racunima. Evo sta se meni desilo,
ja pohadjam treci razred Gimnazije i nema teorije da mogu rijesiti te zadatke jer mi to iz matematike radimo tek u IV razredu. Sad na takmicenju su od 5 zadataka bila 3 sa takvom matematikom. To nije fer!
A upravo te zadatke je sastavljala dr Rajfa Musemovic. Isao sam na zalbu u komisiju i dodjem ja kod nje. Ona mi kaze

      ''Pa sta cu ti ja kad to ne radite!''

A to gradivo uopste nije u nastavnom planu i programu iz fizike i matematike za III razred! :|

To je već druga tema, ali zašto ne i to 'pretresti'. Fizika koju sam ja 'otkrio' na fakultetu prosto ne može da živi bez infinitezimalnog računa (a to je realnost moderne fizike). Korisnik djole sa MathLinksa (Nišlija Đorđe Radičević) je po mom skromnom uvjerenju fenomenalan fizičar, inače moja generacija: u četvrtom razredu je on na MathLinksu primjenjivao matematički aparat (matematička fizika+infinitezimalni račun) kakvim ja kao olimpijac iz matematike nisam ni izbliza raspolagao-prosto, bio mu je neophodan (sad se sjetih da ispričam jedan vic na kraju ovog izlaganja).
Treba provjeriti da li se takav matematski aparat traži na IPhO, ako ne, onda nije uredu, ali ako je odgovor na ovo pitanje potvrdan, onda je u interesu Društva fizičara da na takmičenje odu oni učenici koji barataju tim tehnikama.
Btw, da li učenici u Federaciji uče infinitezimalni račun prije IV razreda? U moje vrijeme nisu, ali ovi reformski procesi su možda nešto promijenili...

A sad vic:

Grupa fizičara i grupa matematičara u vozu: Svaki fizičar je imao jednu kartu, a cijela grupa matematičara samo jednu. Dolazi kondukter, matematičari brzo pobjegnu u WC. Fizičari pokažu svaki svoju kartu, i kondukter ode do WC-a, pokuca i jedan od matematičara mu ispod vrata gurne kartu, ovaj je pogleda
i ode dalje. U povratku se fizičari dogovore da će oni probati istu foru, pa kupe samo jednu kartu, a matematičari nijednu. Kad vide konduktera, fizičari brzo otrče u WC i zaključaju se. Jedan matematicar ode do WC-a, pokuca, uzme im kartu i s drugim matematičarima ode u drugi WC.

Poenta: Fizičari koriste matematičke metode - ali ih ne razumiju.

pa to je izborno takmicenje za IPHO. Ne mozemo mi ocekivat da tih 5 prvoplasiranih tamo ode i zali se na zadatke jer su s takvim racunom xD
nego :) izvinjavam se sto vijest nije prije objavljena ali nisam mogao do kompjutera nikako. 4 clan BIH 2009 IPHO tima je Husremovic Faruk Prva Gimnazija Zenica (bravo majstore) pa eto harune mozes to sad fino sve objavit =)

Takmičenja iz fizike s ovakvim stvarima su stvarno smiješna. To da se mora prolaziti iz entiteta govori o besmislu takmičenja i bolje bi bilo da ga ukinu pa nek bude federalno s kojeg će proći četvero i republično s kojeg će proći jedno, manja je sramota, iako je svakako sramota.
//izvinjavam se na kasnijem javljanju


I posted this some time ago and it got moved.
Prove that polynomial x cannot be a prime finder for all x.

It probably got moved because one of the moderators thought it was beyond the scope of the getting started forum.

Basically a prime finder is a function that when a value x is put in y is a prime.

Also here is the link to the previous thread about this   [url]http://artofproblemsolving.com/Forum/viewtopic.php?t=6611[/url]


I am having trouble understanding how to solve this problem.

Find the value of $ \frac{1}{2!}\plus{}\frac{2}{3!}\plus{}\frac{3}{4!} ... \plus{} \frac{2006}{2007!}$

The answer is $ 1\minus{} \frac{1}{2007!}$

I think this is an MMPC Part I question from two years ago (which is my favorite year  :P )

[hide="solution"] Work with smaller numbers, and you can find some pattern, prove using induction if you want.

Alternatively, as $ \frac{1}{k!}\minus{}\frac{1}{(k\plus{}1)!}\equal{}\frac{(k\plus{}1)!\minus{}k!}{k!(k\plus{}1)!}\equal{}\frac{k}{(k\plus{}1)!}$.

So, $ \frac{1}{2!}\plus{}\frac{2}{3!}\plus{}\frac{3}{4!}...\plus{}\frac{2006}{2007!}\equal{}1\minus{}\frac{1}{2007!}$ as desired.

[/hide]

Yeah that makes sense. Thank you. I had difficulty finding the pattern.

yeah its from MMPC part 1 2007. I'm getting ready for this year so I'm just going through some previous tests.

does anyone have MMPC = Michigan Mathematics Prize Competition / part 2: 2008-09, 2011 problem set?


I have played french horn for a while, this is my 4th year. Does anyone else here play french horn?

yay, im not a solo french horn player!  I think french horn is awesome, although the mouthpiece is small. You have played a little more than me :)

french horn have been renamed Horn since 2007
Why?
they were made in Germany

I played in elementary/middle school band for two years. I quit band class though. Now I just play horn occasionally at home and I think I'm going to start taking lessons soon. My former band director glares at me whenever I see him in the halls (lol)  :roll:

not too many horn players looking in these boards, huh?
well, I'm one :)

my friend plays french horn, but he doesn't have an apos acount

Hm, my sister is thinking about starting french horn, but I have a question:

How much air volume does a french horn player need? I know it's "smaller" than a lot of other brass instruments, but doesn't it unravel to some immensely long length? Does that mean you need a large lung capacity to play it?


Why are angles inscribed in a circle that subtend the same arc necessarily congruent? I've been taking it for granted that they are, but I'd like to see a proof of why.

[hide="Hint"]
Construct an inscribed angle $C$ in circle $D$ that subtends arc $AB$ and the radii $CD, AD$, and $DB$. Note the isosceles triangles formed in your construction and angle chase to prove that $\angle ADB = 2\angle ACB$. Proving that shows that any inscribed angles that subtend the same arc are congruent.

EDIT: Fixed the points $O$ and $D$. thanks bthings.
[/hide]

Got it, thanks! But you might want to keep your $O$ and $D$ straight, I'm pretty sure you used both of those to denote the center of the circle.


what does this mean?

[i][b]f[/b][/i]: R\{1} -> R

$f$ is a function defined for every real number (except $1$) and that returns real numbers.  $\mathbb{R}$ is the set of real numbers, and $\mathbb{R} / \{ x \}$ is the set of real numbers, but without whatever is in the set $\{ x \}$.


I think we should be able to determain what post have been rated quicker than looking at all your post

Actually, you don't get a post rating until 20 of your posts are rated.  :wink:

Some of your posts may actually be rated, but a few people have to rate it before a rating shows up.

Don't worry.  I'm giving you a 3 cube.

@monkeygirl: well, unfortunately, you did spam, and you will have to live with that affect on your rating

I think that what she means is that when you look at your post rating and you want to see how you got that post rating then the only way for you to do that is to look through all your past posts. I also think that there should perhaps be another button in the search function that could filter the posts into rated posts and non-rated posts.
This would also be helpful to spammers, because they could look at the posts that have been rated spam and try to improve their post quality. Otherwise they might not even know that their posts have been rated spam until they go back to the thread and look at their post

[quote="Valentin Vornicu"]There is a feature like this in the works.[/quote]

Sorry to bump an old post, but when is this feature due to be implemented on the boards?

I think these kinds of changes take a while to implement. I also remember hearing that a V2 of the site may be coming out.


trigonometry
geometry
angle bisector
trig identities
Law of Sines
geometry unsolved

[b]Prove that if one angle of a triangle is equal to 120, then the triangle formed by the feet of its angle bisectors is a right angled triangle. [/b]

[hide]
Let $ A$ be the $ 120$ degree angle and let the angle bisectors meet $ BC$, $ CA$, and $ AB$ at $ D$, $ E$, and $ F$ respectively. Then by the Law of Sines to $ \triangle ADC$ and $ \triangle ABC$, we have

$ \frac{AD}{DC}\equal{}\frac{\sin\angle ACD}{\sin 60^{\circ}}\equal{}\frac{\sin\angle ACB}{\sin 120^{\circ}}\equal{}\frac{AB}{BC}$.

Now by the angle bisector theorem, $ \frac{AE}{EC}\equal{}\frac{AB}{BC}\equal{}\frac{AD}{DC}$, so again by the angle bisector theorem, $ DE$ is also the angle bisector of $ \angle ADC$. Similarily, $ DF$ is the angle bisector of $ \angle ADB$. So

$ \angle FDE\equal{}\angle ADF\plus{}\angle ADE\equal{}\frac{1}{2}\angle ADB\plus{}\frac{1}{2}\angle ADC\equal{}90^{\circ}$, so $ \triangle FDE$ is a right angled triangle as desired.
[/hide]

See that AB and AC are external angle bisectors of the angles of 60º made by the angle bisector of A, hence DE and DF are angle bisectors of angle ADB, perpendicular.

Best regards,
sunken rock


inequalities trigonometry induction function

Let $0<x<1$, define a sequence as follows: $x_0=x$ and $x_{n} = \sqrt{\frac{1-\sqrt{1-x_{n-1}^2}}{2}}$ for $n\ge 1$. Prove that $\frac{2x}{\pi}< 2^nx_n < \frac{\pi x}{2}$.

[hide="Solution... I think"]
Let $x_0 = x = \sin \theta$, then it's easy to prove by induction that $x_n = \sin \frac{ \theta }{2^n}$.

Furthermore, $1 - \cos \theta > 0 \implies \theta - \sin \theta > 0 \implies \theta > \sin \theta$, or

$2^n \sin \frac{ \theta}{2^n} < \theta$

We wish to show

$\frac{2 \sin \theta}{\pi} < 2^n \sin \frac{ \theta }{2^n} < \frac{\pi \sin \theta}{2}$

It's easy to show that

$\sin \theta = 2^n \sin \frac{\theta}{2^n} \cos \frac{\theta}{2} \cos \frac{\theta}{4} ... \cos \frac{\theta}{2^n}$

So, rewriting the original inequality, we wish to show

$\frac{2}{\pi} < \frac{1}{\cos \frac{\theta}{2} ... } < \frac{\pi}{2}$, or

$\frac{\pi}{2} > \cos \frac{\theta}{2} \cos \frac{\theta}{4} ... \cos \frac{\theta}{2^n} > \frac{2}{\pi}$

The left hand inequality is obvious and can be strengthened to $1$.

The right hand inequality is more ingenious.  $\cos \frac{\theta}{2} ... \cos \frac{\theta}{2^n} = \frac{\sin \theta}{2^n \sin \frac{\theta}{2^n} }$.  Now, $2^n \sin \frac{\theta}{2^n} < \theta$, so

$\frac{\sin \theta}{2^n \sin \frac{\theta}{2^n} } > \frac{\sin \theta}{\theta}$,

And because of the $\cos$ expansion we know that this is a decreasing function in both $n$ and $\theta$.  The minimum therefore occurs at $\frac{\sin \frac{\pi}{2} }{ \frac{\pi}{2} } = \frac{2}{\pi}$, completing the right hand.

QED.  (Unless I misread the problem and the left hand inequality isn't trivial... )

[/hide]


AMC
USA(J)MO
USAMO
geometry
ARML
AoPS Books

Here is the entire list
[url]http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2009-ua/09-Qual_list.pdf[/url]

Here are the ones from Indiana!  I know like 5 of them!!
NAME            GRADE   SCHOOL                       CITY, STATE, COUNTRY
Lyndon Ji          9    Carmel HS                     Carmel IN USA
Shawn Qian      11    Carmel HS                      Carmel IN USA
David Liang       7     Creekside MS                         Carmel IN USA
David Ramsey   12      Homestead HS                     Ft Wayne IN USA
Richard Ni        10   Park Tudor School                   Indianapolis IN USA
Jeffrey Shen    10     Park Tudor School                   Indianapolis IN USA
Sachin Shinde   11   Terre Haute South Vigo HS    Terre Haute IN USA
Nathan Benjamin 11   Harrison HS                   West Lafayette IN USA
Philip Thomas    12   Harrison HS                   West Lafayette IN USA
Mark Sellke       7    Klondike MS                  West Lafayette IN USA
Yixin Wang       12    West Lafayette HS           West Lafayette IN USA
Justin Ahmann    9    Zionsville Community HS             Zionsville IN USA

Hi, is everyone here taking the exam.  If so, can you guys say how long it took you guys to study and what books you guys used.  I greatly appreciate it and best of luck.  Thanks.

EDIT: I realized that theres already post announcing all the Qualifiers (so this is kind of a repost), but it seems like that one is used for just announcing who qualified.  Can people actually post what did they do qualify?

Hi, yea, so I was very fortunate to qualify the last two years.  Also, just as a quick note, I think I am one of the five you know (pm me and we can talk even more--Im the non-MOSPer from Carmel--thats my hint on who I am lol--anyways I can help you more in detail if you want---and I can help you in school next year).

My biggest advice to you is to work hard.  Like the old adage says, no pain, no gain.

Specifically, I recommend:

1.  Looking through the different forums for different questions posted by various users.  Try to work on the problem, and if you have time, post a solution.  Then go back to the question a couple days later and see (1) if your solution is correct, (2) compare your solution to other people's solutions, so that you can see different ways to work out the same problem.  This is a great website, as personally, I did this and I learned many different tricks, such as balls and urns.

2.  Go to the contest section up top on this website and print out old AMC/AIME problems.  Also, buy some AOPS books, espcially AOPS vol. 1 and vol. 2 because they give a great foundation.  I also advise Intermediate Geometry and Algebra.  You can also work on problems from ARML and other math contests of comparable level.

3.  Go to the user resurrection's dads class (Dr. Ji's class) because he teaches you a lot of great techniques for problem solving...especially geometry and number theory.  You'll also meet a lot of smart students...from the past, many have gone on to elite colleges...and the present, we have like multiple people that have/will make MOSP (or generally referred to as just MOP).

Oh yea so like I said, I think we go to the same school.  So next year, during the school year, I would be willing to help you if you ask me at any time.

And one more thing, I do not know if you heard, but theres a new intermediate thing between AIME/USAMO called the USAJMO.  I think, unless you are not going to be a 10th grader next year (which means I'm mistaking you on who you are), you should first shoot for that in 10th grade and try for the USAMO in 11th.


Can all of us who are not USAMO-ers just pack our bags and assume that we'll never make any positive contribution to mathematics?  Will all the real research only come from the best of the best people from the IMO?  Are we just useless, being semi-good at math but being nothing in comparison to the geniuses?

There are far more people who make useful contributions to mathematics than there are people who go to IMO.  There aren't nearly enough geniuses to keep mathematics research going at any reasonable pace.  Also, the abilities that get one to the USAMO are not a subset of the abilities that make for a good research mathematician and professor (although the intersection is certainly not empty).

So quit feeling sorry for yourself :)

I honestly enjoyed my brief and not too remarkable intervention on Math Olympiads (1st place in my country, but I was already too old to participate in international events). Nevertheless, I consider that the skills you need to be good in math olympiads and the skills that you need to do research are way too different. Nevertheless, I think that a good Math contest's participant has a very good chance to develope researcher's skills. Of course, if you are not good in math olympiads, you still have a good chance to be a good researcher.

On the other hand, look at the Field Medal list of the last, say, fourty years. How many IMO or USAMO participants do you see there. I haven't checked, but I suppose that, even when probably there should be one, two or may be three, most of the good researchers never participated in such contests. They are just different levels.

Thank you for your replies.  I cannot tell you how positive your messages were.  I almost feel like my life has a purpose now.  Now, I'm gonna stop acting so soap operaish, but its really good to hear that everyone else has a chance to discover something in the world of mathematics.

By the way, congradulations Djiminez on first in country.

[quote="RedSickle&Hammer"]By the way, congradulations Djiminez on first in country.[/quote]

Thank you  :blush: ! any way, it was long time ago, 1998! (wow, it starts to feel like too much time  :( )


prove that $n!(m!)^n|(nm)!$.
I have the solution by algebra in the book.
and also my book says the solution by Combinatorics is much easier(but I don't know).
who can post it?

I think the combinatorics solutions is the fact that the number
\[\displaystyle \binom{nm}{n,m,\ldots , m}\]
is an integer .....

[quote="DusT"]\[\displaystyle \binom{nm}{n,m,\ldots , m}\][/quote]
It isn't $\binom{nm+n}{n,m,\ldots , m}\ $?

[quote="Valiowk"]Consider the number of ways of placing $nm$ distinct objects into $n$ identical boxes such that there are exactly $m$ objects in each box.[/quote]
nice,think you.


Jenny has a bag with two of each coin up to a quarter. She picks one coin out. She then picks another. What's the probability she picked out 20 cents or more?

[hide]
First, find the probability that she picked less than 20 cents

All of the pairs of possible coins that add to less than 20 (not counting reverses).
Dime Nickel
Dime Penny
Nickel Penny
Nickel Nickel
Penny Penny

For each of the first three, the probability of selecting those are

$ \frac{2}{8}\cdot\frac{2}{7} \equal{} \frac{1}{14}$

For each of the final two, the probability of selection is

$ \frac{2}{8}\cdot\frac{1}{8} \equal{} \frac{1}{32}$

Adding them all together and then multiplying by 2 to account for selection in the reverse order we get

$ 2[\frac{3}{14} \plus{} \frac{2}{32}] \equal{} \frac{3}{7} \plus{} \frac{1}{8} \equal{} \frac{31}{56}$

Taking the complement, we see that the probability of getting at least 20 cents is $ \frac{25}{56}$
[/hide]


Well, then try to find them by yourself. I am sure you have all the knowledge to do that...

i really don't find a point teaching us that.....
we need to know more fundamental things

She was just mentioning  :D . And i was interested so.....  :)

... so forget about that and concentrate on more important things, like Pardesi said. How is your knowledge of general chemistry?

Then you should develop your skills on that subject. Several problems I have posted on this forum relate to general chemistry (one of them, regarding water's normal pH, remains unanswered).

Ok i will try that  :D . i was conc on organic  :)

i am also weak in physical chem....
i woiuld rate myself 3/10.........post many questions in phy chem and also..tell me th correct approach to the problem..........

[quote="madness"][quote="Carcul"]It should be "Cholesterol". Are you sure this is not a joke? Because proposing a problem like this to students like you seems just that.[/quote]

No she said that as a retrosynthetic problem it was a very good one which had 15 steps in it  :) . I just wanted to know the steps  :D[/quote]

impossible, do you know many chiral carbons cholesterol has?
and in the biosynthesis of cholesterol, it takes 19 steps to convert lanosterol (very similar to cholesterol in structure), which is the precursor of cholesterol, to cholesterol.
and naphthalene does not even remotely resemble cholesterol.

no i am positive  :D  she said 15 steps only  :D

I think she mentioned that cholesterol was converted to naphthalene and from this going via retrosynthetic analysis cholesterol structure was derived.

that is what i meant by:
[quote="madness"]

No she said that as a [u][b]retrosynthetic problem[/b][/u] it was a very good one which had 15 steps in it  :) . I just wanted to know the steps  :D[/quote]

How about posting some of your thoughts on this problem?

As a side note, I also don't think 15 steps are enough: not only the molecule in question have 8 stereogenic centers, but the introduction of that five membered ring would be tricky. Here's an advice: leave this problem and go study more important things.

[quote="Carcul"]How about posting some of your thoughts on this problem?
 Here's an advice: leave this problem and go study more important things.[/quote]

Arent these two lines contradictory?  :huh:

Yes they are. But you can see the second line as just an advice.

There are several problems in this forum that still remain untouched. Look for them.

[quote="chemrock"]I think there has not enough organic problems lately carcul?
can u revive the trend please? Because those problems really ignite our spirits.[/quote]
Rockesh theeeeee :P

[quote="madness"]I cant speak for myself  :D . Ok if i were to compulsarily rate myself then i would give myself probably 4/10 for my level  :)[/quote]
madness is too humble!!! i know him very well as i am his classmate he is the best in our iit classes and scores astonishing totals in chemistry especially. he must be obviously rated 8 or more out of 10 rather than 4 which i would call absurd!!!! :lol:

dont balbber and introducing u to iit classes topper n srinivas  :D  He beats everyone in a 400 mark test by abt 70 marks  :D


I don't quite get this problem.

Let $ f$ and $ g$ be odd functions that are defined for all real numbers.

(a) Is $ f\plus{}g$ odd?
(b) What about $ f \times g$
(c) What about $ f/g$?

My teacher defined it as a function such that f(x) = f(-x).
That would make the answer evident.

EDIT: OOPS! I left out the "-"; thanks, inteluser.

In fact, an even function is a function $ f$ such that $ f(x)\equal{}f(\minus{}x)$.

And odd function is a function $ f$ such that $ f(x)\equal{}\minus{}f(\minus{}x)$.

If we really wanted to, we could make our own odd functions.  $ f(x)\equal{}x^3$ and $ g(x)\equal{}x$. :D

odd functions are functins that satisfy f(x)=-f(-x) for all values of x and:
even functions are functions that satisfy f(x)=f(-x) for all values of x

[quote="ThetaPi"]I don't quite get this problem.

Let $ f$ and $ g$ be odd functions that are defined for all real numbers.

(a) Is $ f \plus{} g$ odd?
(b) What about $ f \times g$
(c) What about $ f/g$?[/quote]

[hide]Let me write the answers:
 a)Yes, it is odd.
 b) It is even
 c) It is even

 You can prove the above very easy, using the given definitions in the post by Inteluser.Question c) needs some remarks, whch consern the definition of $ \frac {f}{g}$ . But the roots of an odd function(here of $ g$ )  are symmetric to the origin, so their quotient is defined on a set symmetric to the origin $ 0$ , so finally $ \frac {f}{g}$ is even.[/hide]

Or, choose 2 odd functions, and then test it.

eitehr way works, but usually you'll haf to prove


[quote="Svejk"]If $a,b,c\geq 0$ prove that:
\[(a+1)^{a-b}\cdot (b+1)^{b-c}\cdot (c+1)^{c-a}\geq (a+1)^{c-a}\cdot (b+1)^{a-b}\cdot (c+1)^{b-c}\]
[/quote]
$(a+1)^{a-b}\cdot (b+1)^{b-c}\cdot (c+1)^{c-a}\geq (a+1)^{c-a}\cdot (b+1)^{a-b}\cdot (c+1)^{b-c}\Leftrightarrow$
$\Leftrightarrow\sum_{cyc}(2a-b-c)\ln(1+a)\geq0,$ which true by  the  rearrangement's inequality or Chebyshov.