ARML
geometry
geometric transformation
rotation

In how many ways can a 2x2x17 hole be filled with 1x1x2 blocks so that the top is even (nothing sticking out) and there are no "holes"?
I have no clue how to do this, but i heard it could be done using recursion. Somebody posted this in the combinatorics section, but nobody seems to respond, so maybe someone in the intermediate forum would be able to explain this.

It seems that recursing on $S_{n}$, the number of ways to fill a $2 \times 2 \times n$ box would work best. It seems to be similar to the two dimensional case, with a $2 \times n$ grid which gives the Fibonacci numbers. The recursion I get is

$S_{n}= 2S_{n-1}+5S_{n-2}$,

where the $2$ corresponds to the number of ways to fill a $2 \times 2 \times 1$ and the $5$ corresponds to the number of ways to fill a $2 \times 2 \times 2$ so that the top and bottom layers cannot be separated (i.e. there is always at least one $1 \times 1 \times 2$ block that is vertical).

[quote="paladin8"][hide="Quoted Recursion"]$S_{n}= 2S_{n-1}+5S_{n-2}$[/hide][/quote]

[hide="I dont think that works exactly."]
What if there are three layers that are inseparable? (Hard to describe without hand gestures, but bottom layer is one horizontal two vertical, then second layer is upper portions of those two verticals, then another two verticals that reach into the third layer.)

[hide="A straight-forward way that I have seen to do this:"]
It gets a li'l nasty, but:

Double recursion. Set the first recursion based, instead of on flat layers, on pieces that are inseparable, flat on one side, and have a $1 \times 2$ sticking out the other end (due to verticals). Set the second recursion based on inseparable pieces with both ends flat. Then see where these pieces can fit together in the overall structure.

I hope I said that right.

Solution credited to [b]besttate[/b].[/hide] [/hide]

Ah, good point. Looks like there's a reason this problem went unsolved...

How do u get the actual recursion relation?

Let $a_{n}$ be the number of ways of filling a $2 \times 2 \times n$ hole with $1 \times 1 \times 2$ blocks.  Consider the leftmost $2 \times 2 \times 1$ end of the hole.  We have three possibilities: we can fill it with 4 bricks (1 way) leaving a $2 \times 2 \times (n-2)$ hole, with 2 bricks (2 ways) leaving a $2 \times 2 \times (n-1)$ hole, or with three bricks (4 ways) leaving a funny-shaped thing.  Thus, $a_{n}= a_{n-2}+2a_{n-1}+4b_{n-1}$.  Now, consider filling the funny-shaped thing.  We can put one block in to make it a $2 \times 2 \times(n-2)$ or we can put in two blocks to make it a shorter funny-shaped thing, so $b_{n-1}= b_{n-2}+a_{n-2}$.  Now, you have a system of two recursions (although you still need some initial conditions), which you can use together for purposes of computation or substitute cleverly to eliminate the $b_{i}$.

Incidentally, when something goes unsolved in Olympiad forums, it would be better to first bump it there, and if still no response, to try in the Pre-Olympiad forums.  This problem is really harder than Intermediate is "supposed" to be.

For those of you not from Chicago, this is a problem that they give to the Chicago ARML D-team, a team reserved for freshmen and sophomores new to ARML.  It's amusing mostly because it is usually only solved by a few people each year, and everyone else is scrambling to try and do it at the last minute.

[hide]
Anyway, for convention sake I will say that the 17 dimension of the hole is down into the ground, let A_n = the numebr of ways of filling a 2x2xn hole, and let B_n be the number of ways of doing it without any horizontal breaks (ie you cannot break it into a 2x2xk and a 2x2x(n-k) block evenly for any k>0).   Ok lets look at A_m, for each filling lets start at the top and go down one foot at a time until we get to the first horizontal break, it must be some where (even if its at the bottom), and above it we have a fillign with no horizontal breaks.  So we find that A_m = the sum from i=1 to n of B_i x A_(m-i).

Ok so what, now we have a sum with two variables? well luckily B_n is really easy to compute.  B_0 = 1 for convention, B_1 = 2 two horizontal oriented two ways, B_2 = 1 four oriented vertically, B_n, n>2 = 4 the bricks sort of zig-zag up (hard to describe) and there are 4 rotations.

So anyway, the first few terms you need to compute yourself, but then you get:
A_m = 2A_(m-1) + A_(m-2) + 4A(m-3) + ...+ 4A_0
A_(m-1) = 2A_(m-2) +A(m-3) + 4A_(m-4) + ...+4A_0

A_m - A_(m-1) = 2A_(m-1) - A_(m-2) +3A_(m-3)

A_m = 3A_(m-1) - A_(m-2) +3A_(m-3)

Now we have a nice recursion to work with, all thats left is finding the first few values. We can easily see that A_0 = 1, A_1 = 2, A_2 = 5.  Now plug and chug and you get some ridiculous number that I dont want to compute.[/hide]

[quote="Kamil Witek"]wat's the funny shaped thing?[/quote]

I think the explanation was fairly clear: cover the $2 \times 2 \times 1$ end portion of the hole with 3 bricks.  (There are 4 ways to do this and they are all rotations of each other.)  The resulting shape is what I called a "funny shaped thing."  You have to actually think about what this looks like.  It is also the same shape (up to overall length) that you get by plunking a single $1 \times 1 \times 2$ block into the hole.  (This is approximately equivalent to nateharman's "sort of zig-zag up (hard to describe)," although I think my solution is nicer  :wink: )

With JBL's notation, $ a_{n}=3a_{n-1}+3a_{n-2}-a_{n-3}$, $ n\geq 4$. What are the initial conditions? $ a_{1}$ is obviously 2.

$ a_{0}= 1$, $ a_{1}= 2$, $ b_{1}= 1$ is all you need to get $ a_{2}$ and you can go from there.  (Nateharman incorrectly wrote that $ a_{2}= 5$, but it's not too hard to count directly to see that it's actually larger.)

$ a_{0}$ is the number of ways to fill a $ 2 \times 2 \times 0$ hole with $ 1 \times 1 \times 2$ bricks.  You can do this in exactly 1 way, using 0 bricks.  This is mildly risky (sometimes the way you define things recurrences will break going backwards) but usually valid -- in this case, it makes perfect sense in the recursion.  The other thing to do is just count directly that $ a_{2}= 9$.  (I think this is easiest to do by the same method that I arrived at the recurrence, i.e. consider how you can cover a fixed face of the $ 2\times2\times2$ cube.)


Are there any requirements in grad school regarding knowing other languages besides english in usa?

I have read that the United States is the ONLY country in the world in which it is possible to earn a Ph.D. degree without learning another language. That's pretty pathetic. Really good graduate programs in math (e.g., Princeton's) expect Ph.D. candidates to show ability to read another language for math research, but conversational ability is widely ignored here. If you already know another language besides English, you are way ahead of most American graduate students.

So which language would be recommended to learn in order to be able to " read another language for math research"?
German? French? Chinese? Latin?

Most math graduate schools require you to pass exams in two of {French, German, Russian}. But the exams are pretty easy: you usually have to translate a passage from a math book or paper, and generally you're allowed to use a dictionary. Keep in mind that it's a lot easier to read a math book in a foreign language than practically anything else since most of the math words are cognates!

Some graduate programs specify the languages. The most often specified languages are German, French, or Russian. I'm not aware of any graduate program yet that lists Chinese (my children's other language), but perhaps someday that will be a possibility. These days most mathematical journal articles are published in English, by a huge margin, but there are still a few French-language-only journals published in France.

Choosing languages is right now a question I have to deal with. So here are a few more questions.
I haven't had any time too look into the whole application process but I assume that I'll have to write the languages I know in some paper which is sent to them (the univ.) and they make the calls based on that (of course, this is before the exam ComplexZeta mentioned). Is there a minimum amount of time you have had to read a language for them to accept you (as having enough proficiency in that language and staging you to the next level)? I have read German in 3.5 years in middle school, do you think they would accept this as enough? I am also reading a course in French which my teacher compares to 1.5 years in High School, enough?

My second question is this : Assume that they say "okay 3.5yrs in MS and 1.5yrs in HS is enough" and I get accepted to the university. But later (I assume the tests are taken there?) fail one or both the exams. What consequences will this give? Are they going to kick me out of their graduate program or will I be forced to take German a French courses there at campus?

Thank you!

kahsi, as near as I can tell, prior language proficiency plays zero role in the admissions process to any U.S. graduate program in math that I know about - your middle school German will make no difference either way. They mostly just sort of assume you'll figure out how to pass the required exams. There may be courses to help, but I'm guessing most people don't take them. At my graduate school (nearly 30 years ago) I passed exams in German and French; I'd had some high school and college German, which helped, but I had exactly zero background in French. The Latin/Greek big words are the same; the symbols are the same; you practice a little bit with some papers so you get used to how a mathematical writer would say things like "suppose that" or "let." Where I was, one was allowed to use a dictionary. (I had an ordinary French/English dictionary - there's no need for a technical dictionary, becuase the technical words are all the same anyway.) It's a little cheap - there's not a chance in the world that I could have read a French newspaper, never mind a piece of literature. But mathematical papers are different.

I remember stumbling a bit on the German exam, faced with a piece of linear algebra, wondering what the word "Körper" (sp?) could mean, but the mathematical context was such that the only reasonable meaning was "field" (or maybe division ring, but that seemed unlikely.)

Indeed, it sounds cheap.
From reading your text I get the feeling that the exam isn't taken directly (you mentioned that people could take courses first), then when is it taken? Also, since it is taken a bit late I assume they wont kick you out when failing the exam, correct?

I received my math Ph.D. from MIT.  For what it's worth, here is their current policy (which I got off their website):

[quote="MIT"]While there is no Institute Language Requirement, doctoral candidates in the Department of Mathematics (both Pure Mathematics and Applied Mathematics) have to demonstrate a mathematical reading knowledge in at least one of the following languages: French, German, Italian, Russian, Spanish or Chinese. This proficiency can be demonstrated in any of the following ways:

   1. native fluency
   2. two years of the language in high school with the grade of C or better
   3. one year of the language in college with the grade of C or better
   4. one semester of the language courses being given at MIT with the grade of C or better
   5. taking the language examination given at MIT
[/quote]
I satisfied this requirement on my first day there by walking into the department office and telling them that I took 4 years of French in high school.  They said "OK" and checked off on my file that I had satisfied the language requirement.  Note that I did not have any documentation for this; they simply took my word for it.

As it turned out, since I ended up studying algebraic geometry, I did have to read a lot of papers in French, in particular some of [i]Éléments de géométrie algébrique[/i] by Grothendieck.

will it help me to take 4yrs latin+ 4yrs french? (im in 8th grade)

im taking them no matter what anyways
jorian

Taking Latin and French will certainly be educational for you, and will likely increase your recognition of English words from those languages.


2 same ball are connected by a massless unstretchable string length l. Put them on a frictionless horizontal plane as showed in the picture. At t=0 one of these balls get a vertical speed of V0. Find the value of V0 so that the under ball is alway at the plane and the string is always straight.

[code]                   ^ V0
                   |
                   |
     O-------------O
===============================[/code]

There are no dissipative forves present. The centre of mass of these two bodies will move in a straight line vertically upwards. And the critical moment when both balls will tend to bounce is at the top and at this postion, $ \frac{m(v')^2}{R} \le (m\plus{}m)g$ where $ r\equal{}\frac{l}{2}$. The velocity of the ball at top in left direction at this critical moment will be equal to velocity of the lower ball in right direction.
And this velocity, $ v'\equal{}\frac{v_0}{2}$.
Now solving further will give you the correct answer.

[quote="Rajiv"]There are no dissipative forves present. The centre of mass of these two bodies will move in a straight line vertically upwards. And the critical moment when both balls will tend to bounce is at the top and at this postion, $ \frac {m(v')^2}{R} \le (m \plus{} m)g$ where $ r \equal{} \frac {l}{2}$. The velocity of the ball at top in left direction at this critical moment will be equal to velocity of the lower ball in right direction.
And this velocity, $ v' \equal{} \frac {v_0}{2}$.
Now solving further will give you the correct answer.[/quote]
What worries me is the feeling that this assumption is wrong. I think, the right ball moves vertically upwards and at the top, when the other ball is right underneath, exerts a jerk (read "imparts momentum") and that is the critical moment. You need to find out the velocity by means of energy conservation and momentum conservation....and that may involve a little of integration work for the momentum part but eventually the question is quite do-able!


Let $S$ be the set of all rational numbers $r$, $0<r<1$, that have a repeating decimal expansion in the form \[0.abcabcabc\ldots=0.\overline{abc},\] where the digits $a$, $b$, and $c$ are not necessarily distinct. To write the elements of $S$ as fractions in lowest terms, how many different numerators are required?

[hide]Notice that $0.\overline{abc}=\frac{abc}{999}$. The fraction is automatically in lowest terms if $abc$ is not divisible by $3$ or $37$. Now, we take note that the number $\frac{3}{999}$ has the same numerator as $\frac{1}{999}$when reduced to lowest terms, and similarly for other multiples of $3$ and $37$.

There are $333$ multiples of $3$, and $27$ multiples of $37$, so we subtract to get $999-333-27=639$. However, we undercounted because $9$ numbers are multiples of both $3$ and $37$, which we subtracted twice.

Now, notice that the $12$ numbers divisible by $81$ have now been counted, though they provide different numerators. Take $\frac{81}{999}=\frac{3}{37}$. Thus, the answer is $639+9+12=\boxed{660}$.[/hide]

[quote="Farenhajt"][hide]$0.\overline{abc}={abc\over 999}=\frac{abc}{3^3\cdot 37}$, hence we'll need $(3+1)(1+1)-1=7$ numerators, which is number of divisors of $999$ excluding $1$.[/hide][/quote]Which one of us misread?  :huh:

[hide="solution"]pretty obvious that it has to be $\frac{n}{999}$, note that if the 3 digit number has a factor of $3$ or $37$, then it can reduce, and we would have over counted, so use PIE, $f(n)=\left\lfloor\frac{998}{n}\right\rfloor$, so our answer is $998-f(3)-f(37)+f(3*37)=998-332-8+2=\boxed{660}$[/hide]

I think I'm missing something... What about numbers like $.\overline{081}=\frac{81}{999}=\frac{3}{37}$?  It seems to me that numerators that are multiples of 3 are possible.

Well, we want numerators that start out with a multiple of 3, that way it reduces when divided by 999.  So, yeah, in the case you provided, it can end up as a multiple of 3, but that is irrelevant.

Altheman, I did it the same as you, although wouldn't (using your notation) $f(37) = 26$ and $f(3 * 37) = 8$?  I got those numbers, and got a solution of
[hide] 648 [/hide]

So, we are subtracting the multiples of 3 and 37 because they will reduce to numerators we already have.

Is that correct?

I think matt276eagles has a point. The example he gives of $\frac{81}{999} = \frac{3}{37}$ shows that $3$ CAN be a numerator, but is discounted. Similarly, $\frac{k \cdot 81}{999} = \frac{3k}{37}$ so actually quite a few of the multiples of $3$ can be numerators...

I think we'll have to add back $12$ for $k = 1,2,\ldots,12$ to get $672$...

[quote="4everwise"]That makes sense, but I just checked the answer to this problem and it was still the original one.  :huh:[/quote]

Ah, reading over your initial solution again, you did add $12$ at the end to account for the numbers I mentioned, but I couldn't really understand your reasoning there... Altheman's solution seems a little off (pointed out earlier with the wrong values for $f$) but he accidentally gets the same answer, I guess...

Yeah, I've edited the solution, so it makes sense now.  :)

i don't know what i was thinking...i don't see how i got the right answer then, i did 8/3=2. ..., but i should have multiplied by 3? that is weird, oh well

I got $7$, but I realize I'm wrong. 7 is assuming that your numerator is always going to be 1, but obviously it isn't true for all cases.

a Quick Question. Noticed you wrote $ \phi(999) $. What is that notation , what does it mean what does it do? I've seen $ \phi(x) $ couple times but assumed it was function (ie $f(x)$) or placeholder variable (ie $\theta$). Obviously its a symbol that illustrates some sort of.. I'm not sure, but $ \phi(999) $. is 648 so... please help Thanks.

[url]http://www.artofproblemsolving.com/Wiki/index.php/Euler's_totient_function[/url]

it's called Euler's totient function

[quote="joeislittle"][hide="slightly different approach to this old question"]We first count the number of possible numerators when $ 999$ is the denominator.  This gives $ \phi(999)$ possibilities.  Note that if our denominator is a multiple of both $ 3$ and $ 37$, our numerator was covered in the first case.  If our denominator is a power of $ 3$, then our numerator would have to be a multiple of $ 37$ to avoid being included in the first case.  This is impossible.  If our denominator is $ 37$, our numerator would have to be a multiple of $ 3$ to avoid being included in the first case.  This gives $ \frac{36}{3}\equal{}12$ more possible numerators.  Therefore the total number of possible numerators is $ \phi(999)\plus{}12\equal{}660$.[/hide][/quote]

can someone please reword this; I dont understand why the answer is not $998-\phi{(999)} = 998 - 999\dfrac{998}{999}\cdot\dfrac{2}{3} = 350$.

Could I get a little help?  I'm not quite seeing why it's not $648$ by PIE with number of factors of 3+number of factors of 37-number of factors of 111.  The solution says you add back the number of factors of 81, but I don't see why- don't they just reduce to a numerator we already counted?

We excluded all possible numerators that are a multiple of 3. (ie 3, 6, 9, 27, 663 are all excluded) We find out that if the numerator is a multiple of 81, then, when simplified, the resulting fraction will have a numerator of multiple 3, so we must "add back" those multiples of 81.

Say our unreduced is in the form $\frac{81x}{999}$, where $x$ is an integer and $x > 0$. Our fraction will simplify to $\frac{3x}{37}$. See, our numerator is a multiple of $3$. This generates a couple more cases, $12$ to be exact

By the way, because $0<r<1$, is the $r \neq 0.999999... =1$ ?
If so, the correct answer is
$998- \Bigg \lfloor {\frac {998}{3}} \Bigg\rfloor - \Bigg \lfloor {\frac {998}{37}} \Bigg\rfloor +\Bigg \lfloor {\frac {998}{3 \times37}} \Bigg\rfloor +12 =660$ ...?

@above not the same. this was easier and much harder to silly on

Must be between $1$ and $999$, inclusive.

Case 1: Relatively prime to $999$.
Then there are $3^2\cdot 2\cdot 36=648$ ways.

Case 2: Divisible by $37$.
Then less than or equal to $27$, contradiction.

Case 3: Divisible by $3$.
Then less than or equal to $37$, so $12$ ways.

Answer is $648+12=\boxed{660}$.


calculus
derivative
calculus computations

$ sin(arcsin(a))\equal{}a$
so $ g(a)\equal{}a$ and $ g'(a)\equal{}1$

the part about where it is defined refers to $ \minus{}1<a<1$ because arcsine isn't defined if a is greater than 1 or less than negative 1.


[quote="dondigo"]Find the maximum value of: $\sqrt{x+7}+\sqrt{11-x}$,    $x\in\mathbb{R}$[/quote][hide="Answer"]Let $f(x) = \sqrt{x+7}+\sqrt{11-x}$ then $f'(x) = \frac{1}{2(x+7)}-\frac{1}{2(11-x)}$ so $2x+14 = 22-2x \Rightarrow 4x=8$ So $x= 2$ this makes the expression $\sqrt{x+7}+\sqrt{11-x}=\boxed{6}$[/hide]
Edit: Look at that, same exact post time yet t0rajir0u beat me to it  :(

[quote="Hokkage"] then $f'(x) = \frac{1}{2(x+7)}-\frac{1}{2(11-x)}$[/quote]

You forgot the square roots and I forgot to simplify my solution to $f(2) = 6$  :D

[hide]Since the square root of a real number must be positive, RMS-AM inequality can be used.

$\frac{f(x)}{2}\leq \sqrt{\frac{(x+7)+(11-x)}{2}}=3\Rightarrow f(x)\leq \boxed{6}$.

Equality holds when $x+7=11-x$, which is when $x=2$.

(RMS-AM: $a,b\in \mathbb{R}^+ \Rightarrow \sqrt{ \frac{a^2+b^2}{2}}\geq \frac{a+b}{2}$.)[/hide]

[quote="t0rajir0u"]You forgot the square roots[/quote]Whoops :roll: and I'm supposed to be a calc student.  :P
Not going to bother editing though. Your solution's good.

[quote="matt276eagles"]What's the RMS-AM inequality?[/quote]

RMS stands for root mean square; given a sequence $x_k$, it is given by:

$\sqrt{\frac{x_1^2+x_2^2+\cdots +x_n^2}{n}}$

When the $x_i$ are nonnegative real numbers, RMS-AM inequality says the root mean square of the terms of a sequence is always greater than or equal to the arithmetic mean. There is equality if and only if all of the $x_k$ are equal.

[quote="matt276eagles"]What's the RMS-AM inequality?[/quote]By use of the almighty Google, [quote="DPatrick (19:37:25)"]The RMS-AM inequality tells us that the root-mean square (quadratic mean) of a set of real numbers is at least as large as the arithmetic mean of that set. For three variables this looks like $\sqrt{\frac{a^2+b^2+c^2}{3}} \ge \frac{a+b+c}{3}$[/quote]Link: [url]http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=74[/url]

[quote="MysticTerminator"][quote="pkothari13"]Ah, that's a cool way to solve it.  Would there be a way to find the maximum value of $\sqrt{2x-3} + \sqrt{7-4x}$ (I just chose random numbers there), where the RMS won't turn out as nicely (without using calculus)?[/quote]

$\sqrt{2x-3} + \sqrt{7-4x}$ $=$ $\sqrt{2x - 3} + \sqrt{\frac{7}{4}-x} + \sqrt{\frac{7}{4} - x}$ $\le$ $3 \sqrt{\frac{\sqrt{2x - 3}^2 + \sqrt{\frac{7}{4}-x}^2 + \sqrt{\frac{7}{4} - x}^2}{3}}$ $=$ $\sqrt{3 \cdot ((2x-3) + (\frac{7}{4} - x) + (\frac{7}{4} - x))}$ $=$ $\sqrt{3 \cdot \frac{1}{2}}$ $=$ $\sqrt{\frac{3}{2}}$.[/quote]

...............

WOW.  :cool:


Here's how the game works.  Each new player starts with 50 points.  Every time you post, you may bring one person's rating down, and one person's rating up.  You may not select yourself.  Hmmm, did I forget anything?

ernie- 50
[color=red]mz94- 49[/color]
[color=green]stevenmeow- 51[/color]

I just chose random people. :D

ernie- 50
mz94- [color=green]50[/color]
stevenmeow- 51
tinytim-[color=red]50-1=49[/color]

[quote="ernie"] You may not select yourself.[/quote]

Ahem... (tinytim, "Hem, hem")

ernie - 50
mz94 - [color=red]49[/color]
stevenmeow - [color=green]52[/color]
tinytim - 49
Yongyi781 - 50

ernie - 50
[color=green]mz94 - 50[/color]
stevenmeow - 52
[color=red]tinytim - 48[/color]
Yongyi781 - 50

gosh, why do so many people hate mz94?

darn, I am last. :(

[quote="tinytim"]ernie - 50
[color=green]mz94 - 50[/color]
stevenmeow - 52
[color=red]tinytim - 48[/color]
Yongyi781 - 50

gosh, why do so many people hate mz94?

[b]darn, I am last. [/b]:([/quote]

Is that sarcasm?

ernie - 50
[color=red]mz94 - 49[/color]
stevenmeow - 52
[color=green]tinytim - 49[/color]
Yongyi781 - 50

ernie - 50
[color=green]mz94 - 50[/color]
stevenmeow - 52
[color=red]tinytim - 48[/color]
Yongyi781 - 50[/quote]

BLAH

Kamikazee

1=2 - 0
ernie - 50
mz94 - 50
stevenmeow - 52
tinytim - 0
Yongyi781 - 50

ernie - 50
mz94 - 50
stevenmeow - 52
tinytim - 47
Yongyi781 - 50
ternary0210-50

ernie - 50
mz94 - 50
stevenmeow - 52
tinytim - 46
Yongyi781 - 50
ternary0210-50

ernie - 50
mz94 - 50
stevenmeow - 52
tinytim - 45
Yongyi781 - 50
ternary0210-50

[color=green]ernie - 51[/color]
mz94 - 50
stevenmeow - 52
tinytim - 45
Yongyi781 - 50
[color=red]ternary0210-49[/color]
evilhamster - 50

ernie - 51
mz94 - 50
[color=red]stevenmeow - 51[/color]
tinytim - 45
Yongyi781 - 50
ternary0210-49
[color=green]evilhamster - 51[/color]

ernie - 51
mz94 - 50
stevenmeow - 51
[color=red]tinytim - 44 [/color] (picking on the lowest one)
Yongyi781 - 50
[color=green]ternary0210-50 [/color]
evilhamster - 51
dojo-50

ernie - 51
mz94 - 50
stevenmeow - 51
tinytim - 43
Yongyi781 - 51
ternary0210-50
evilhamster - 51
dojo-50

ernie - 51
mz94 - 50
stevenmeow - 51
[color=red]tinytim - 42 [/color]
[color=green]Yongyi781 - 52[/color]
ternary0210-50
evilhamster - 51
dojo-50
shentang-50

[color=red]ernie - 50[/color]
mz94 - 50
stevenmeow - 51
[color=green]tinytim - 43 [/color]
Yongyi781 - 52
ternary0210-50
evilhamster - 51
dojo-50
shentang-50

nobody likes tinytim?

kill ernie for attempting to kill em

ernie - 50
[color=red]mz94 - 49[/color]
stevenmeow - 51
tinytim - 43
[color=green]Yongyi781 - 53[/color]
ternary0210-50
evilhamster - 51
dojo-50
shentang-50

FAIL mz.

ernie - 50
[color=red]mz94 - 48[/color]
stevenmeow - 51
tinytim - 43
Yongyi781 - 53
ternary0210-50
[color=green]evilhamster - 52[/color]
dojo-50
shentang-50

Fail mz.

ernie - 50
mz94 - 48
stevenmeow - 52
tinytim - 43
Yongyi781 - 53
ternary0210-50
evilhamster - 51
dojo-50
shentang-50
lotsofmath-50

ernie - 50
[color=red]mz94 - 47[/color]
stevenmeow - 52
[color=green]tinytim - 44[/color]
Yongyi781 - 53
ternary0210-50
evilhamster - 51
dojo-50
shentang-50
lotsofmath-50

Fail mz.
Fail tinytim.

:P :P :P

Yay new page!

ernie - 50
[color=red]mz94 - 46[/color]
stevenmeow - 52
[color=green]tinytim - 44[/color]
Yongyi781 - 53
ternary0210-50
evilhamster - 51
dojo-50
shentang-50
lotsofmath-50

Fail mz.
Fail tinytim.

:P :P :P

ernie - 50
mz94 - 46
stevenmeow - 52
tinytim - 44
Yongyi781 - 53
[color=red]ternary0210-49 -.- die[/color]
evilhamster - 51
dojo-50
shentang-50
lotsofmath-50
[color=green]Anaxerzia - 51[/color]

gosh, this is so depressing... Why does everyone hate me?

ernie - 50[color=green]
mz94 - 47[/color]
stevenmeow - 52
[color=red]tinytim - 43[/color]
Yongyi781 - 53
ternary0210-49
evilhamster - 51
dojo-50
shentang-50
lotsofmath-50
Anaxerzia - 51

[hide="List"]Yongyi781 - 53
stevenmeow - 52
Anaxerzia - 51
evilhamster - 51
Dojo-50
ernie - 50
lotsofmath-50
shentang-50
ternary0210-49
[color=red]mz94 - 46[/color]
[color=green]tinytim - 44[/color][/hide]

Yongyi781 - 53
stevenmeow - 52
Anaxerzia - 51
evilhamster - 51
Dojo-50
ernie - 50
lotsofmath-50
shentang-50
ternary0210-49
[color=red]mz94 - 45[/color]
[color=green]tinytim - 45[/color]

[color=green]Yongyi781 - 54[/color]
stevenmeow - 52
Anaxerzia - 51
evilhamster - 51
Dojo-50
ernie - 50
lotsofmath-50
shentang-50
ternary0210-49
mz94 - 45
[color=red]tinytim - 44[/color]

Yongyi781 - 54
stevenmeow - 52
Anaxerzia - 51
evilhamster - 51
Dojo-50
ernie - 50
lotsofmath-50
shentang-50
ternary0210-49
[color=green]tinytim - 45[/color]
[color=red]mz94 - 44[/color]

Fail tinytim.
Fail mz.

Yongyi781 - 54
stevenmeow - 52
Anaxerzia - 51
evilhamster - 51
Dojo-50
ernie - 50
lotsofmath-50
shentang-50
ternary0210-49
tinytim - 46
mz94 - 44

Yongyi781 - 54
stevenmeow - 52
Anaxerzia - 51
[color=red]evilhamster - 50 [/color]
Dojo-50
ernie - 50
lotsofmath-50
shentang-50
ternary0210-49
[color=green]tinytim - 47 [/color]
mz94 - 44

[color=red]Yongyi781 - 53[/color]
stevenmeow - 52
Anaxerzia - 51
evilhamster - 50
Dojo-50
ernie - 50
lotsofmath-50
shentang-50
[color=green]ternary0210-50[/color]
tinytim - 47
mz94 - 44

I feel like dragging everyone down.

Yongyi781 - 53
stevenmeow - 52
Anaxerzia - 51
evilhamster - 50
Dojo-50
ernie - 50
lotsofmath-50
shentang-50
[color=red]ternary0210-49[/color]
tinytim - 47
[color=green]mz94 - 45[/color]
xpmath - 50
I just chose random people :P

Yongyi781 - 53
stevenmeow - 52
Anaxerzia - 51
[color=green]xpmath - 51[/color]
evilhamster - 50
Dojo-50
ernie - 50
lotsofmath-50
shentang-50
ternary0210-49
tinytim - 47
[color=red]mz94 - 44[/color]

Yongyi781 - 53
stevenmeow - 52
Anaxerzia - 51
xpmath - 51
[color=green]shentang-51[/color]
evilhamster - 50
Dojo-50
ernie - 50
lotsofmath-50
ternary0210-49
tinytim - 47
[color=red]mz94 - 44[/color]

Yongyi781 - 53
stevenmeow - 52
Anaxerzia - 51
xpmath - 51
shentang-51
dnkywin - 50
evilhamster - 50
Dojo-50
ernie - 50
lotsofmath-50
ternary0210-49
[color=green]tinytim - 48[/color]
[color=red]mz94 - 43[/color]

Let's kill mz!

Yongyi781 - 53
stevenmeow - 52
Anaxerzia - 51
xpmath - 51
shentang-51
dnkywin - 50
evilhamster - 50
Dojo-50
ernie - 50
lotsofmath-50
ternary0210-49
[color=green]tinytim - 49[/color]
[color=red]mz94 - 42[/color]

Yongyi781 - 53
stevenmeow - 52
Anaxerzia - 51
[color=green]dnkywin - 51[/color]
shentang-51
xpmath - 51
evilhamster - 50
Dojo-50
ernie - 50
lotsofmath-50
ternary0210-49
tinytim - 49
[color=red]mz94 - 41[/color]

[color=green]Yongyi781 - 54[/color]
stevenmeow - 52
Anaxerzia - 51
dnkywin - 51
shentang-51
xpmath - 51
evilhamster - 50
Dojo-50
ernie - 50
lotsofmath-50
ternary0210-49
tinytim - 49
[color=red]mz94 - 40[/color]

Failed game.

Yongyi781 - 54
[color=green]stevenmeow - 53[/color]
Anaxerzia - 51
dnkywin - 51
shentang-51
xpmath - 51
evilhamster - 50
Dojo-50
ernie - 50
lotsofmath-50
ternary0210-49
tinytim - 49
[color=red]mz94 - 39[/color]

stevenmeow - 54
Yongyi781 - 53
Anaxerzia - 51
dnkywin - 51
shentang-51
xpmath - 51
evilhamster - 50
Dojo-50
ernie - 50
lotsofmath-50
ternary0210-49
tinytim - 49
mz94 - 39


algebra
polynomial
number theory
superior algebra
superior algebra unsolved

Let $ \omega\equal{}e^{2\pi/n}$.
Which prime number in $ \mathbb{Z}$ still a prime element in $ \mathbb{Z}[\omega]$?

I think you meant $ w \equal{} e^{\frac {2 \pi i}{n}}$, a primitive $ n$-th root of unity.

It's a standard fact that for a rational prime $ p$, $ p$ ramifies if and only if $ p$ divides $ n$, and for $ p$ not dividing $ n$, the residue field degree (of any prime above $ p$ in $ \mathbb{Z}[w]$) is the smallest $ f$ such that $ p^f \equal{} 1$ (mod $ n$).  If you want $ p$ to remain a prime, you need it to be unramified, i.e. $ f \equal{} phi(n)$.  So the rational primes that remain a prime are those $ p$ that are primitive roots mod $ n$.

Hi, thank you for your explanation.
Can you give me a reference that contains the result you mentioned above?

One more silly question.
If $ p$ prime in $ \mathbb{Z}[\omega]$ and $ p\mid \sum_{k\equal{}0}^{n\minus{}1}a_k\omega^{k}$. Does this implies that $ p\mid a_k$ for all $ k$?

No. After all, $ p|1\plus{}\omega\plus{}\omega^2\plus{}\cdots\plus{}\omega^{n\minus{}1}$, and it doesn't divide $ 1$.

[quote="barra"]Hi, thank you for your explanation.
Can you give me a reference that contains the result you mentioned above?
[/quote]

Cassels and Frohlich's Algebraic Number Theory has a small, nice chapter on Cyclotomic Fields (but be warned, this book is out of print.  But you may ask William Stein for help in obtaining this), which is based on B.J.Birch's account.

Algebraic Number Theory by Jürgen Neukirch also has this elementary result, though I personally find Neukirch expositions a little too lengthy in most cases (not just this book, but his cohomology of number fields as well).  But that's just my opinion.

I believe the famed Number Fields by Daniel Marcus also has this result, but I never read this book (shame).  The only intro to algebraic number theory/class field theory books I read from cover to cover are the above two.

Serge Lang should have this too.  I mean, just about any introductory course in algebraic number theory will have this.  Afterall, the proof only uses CRT, and an easy fact about extensions of finite fields.

In fact I can just give you a sketchy proof where you can fill in all the details by yourself as an exercise:

You just want to understand how the cyclotomic polynomial over $ \mathbb{Q}$ factors mod $ p$, as this determines how $ p$ ramifies in $ \mathbb{Z}[w]$.  As everything is galois, every place over $ p$ will have the same ramification index and degree.  If $ p$ divides $ n$, then $ p$ ramifies (exercise for you!).  If $ p$ doesn't divide $ n$, then the cyclotomic polynomial divides $ X^n \minus{} 1$, which is separable even over $ \mathbb{F}_p$.  So the n-th roots of unity in $ \mathbb{Z}[w]$ map bijectively to $ n$-th roots of unity mod $ p$.  And smallest field extension of $ \mathbb{F}_p$ containing $ n\minus{}th$ roots of unity is $ \mathbb{F}_{p^f}$ where $ f$ is the order of $ p$ mod $ n$.  And this coincides with the splitting field of $ X^n \minus{} 1$, so this is the residue field for any prime lying above $ p$.


geometry
3D geometry
sphere
geometric transformation
reflection
trigonometry

A point source emits light(spherically symmetrically) and is located along the axis of a cone (with vertex angle $\theta$,at a distance d from the tip.The inside surface of the cone is reflective.A receiver is located inside the cone.The receiver consists of all points inside the cone that are a distance r from the tip.What fraction of the light emitted by the source hits the receiver?

any ideea?
is there an easier way than considering the tangents from the piece of sphere (of radius r)in the cone that ,by intersecting the cone axis,give us many posibilities depending on the number of reflections?

But there is only one way that you don't need the number of reflections.Think about that... :)

could you tell me what is this way consisting of?

Don't reflect the light from the cone.Just draw a tangential line to the circle.You can see in the attachment,I drew the full circle.

There isn't $\theta$ in the answer. :( The answer is different.
$\frac{1}{2}(1-\frac{(d^{2}-r^{2})^\frac{1}{2}}{d})$

By the way is $\theta$ the opening angle of the cone?

but shouldn't there be theta in the answer if it's given?why did you reached this way of resolving namiqfizik?what about you thaakisfox?

Of course $\theta$ should be in the answer.

In the solution of the problem we doesn't need $\theta$.If you want I will post the solution.

is the official sollution in which we don't need theta?

Here is the solution.

$A_{0}=4\pi R^{2}$

$dA=2\pi y dy =2\pi Rsin\theta Rd\theta$

$dA=2\pi R^{2}sin\theta d\theta$
We integrate both sides of equation and get the following:

$A=2\pi R^{2}(1-cos\beta)$
$\beta$ is the angle between the tangent line and the line connecting the center of receiver and the point source.

$cos\beta=\frac{\sqrt{d^{2}-r^{2}}}{d}$

$k=\frac{A}{A_{0}}=\frac{1}{2}(1-\frac{\sqrt{d^{2}-r^{2}}}{d})$

$A_{0}$ is the light emitted,and $A$ is thelight recieved area  in three-dimensional plane

i also got your answer namiqfizik if that was the way of solving it.but having theta and not using it means you're wrong if it's not the official sollution :) .why did you draw the tangent anyway?how does this geometrically explain that any ray reflected goes to the receiver?

But the official solution is also as mine. :) If you solved the problem,you must know that everything is Ok in the solution.And reflection of rays is just of a mirror rule.I combined the rays which are reflected.


geometry
3D geometry
rotation
combinatorics unsolved
combinatorics

On Rubik's cube, is there a single "move" (where a move is a combination of simple moves, like left "up" twice, rotate front face 90 deg then left "down" twice, etc.) that when applied again and again will cycle through every possible position which the cube can be in? thus, no matter where we start, applying this "move" will solve the cube every time.

If so, the group of permutations of the cube would be cyclic and thus abelian.  However, it's easy to find pairs of moves which don't commute.


geometry
rectangle
rotation
vector
linear algebra
matrix
analytic geometry

this question is realy nice but its solution is realy tough in my document .
with tones of linear algebra lemma. by the way please help me :(  :?

A figure is made out of unit squares joined along their sides. It has the property that if the squares of an m x n rectangle are filled with real numbers with positive sum, then the figure can be placed over the rectangle (possibly after being rotated, but with each square of the figure coinciding with a square of the rectangle) so that the sum of the numbers under each square is positive. Prove that a number of copies of the figure can be placed over an m x n rectangle so that each square of the rectangle is covered by the same number of figures.

In what follows, all vectors are column vectors. Given a vector $s$, the notation $s\ge 0$ means that all the entries of $s$ are $\ge 0$, and similarly for $>0,\le 0,<0$.

Note that the sets of squares we're concerned with are contained in an $m\times n$ rectangle. To each such set we can associate its characteristic vector as a subset of the $m\times n$ rectangle. Let $A$ be the $t\times k$ matrix having the characteristic vectors of the pieces as its columns, where $t=mn$ and $k$ is the number of different positions in which the piece can be placed inside the rectangle, and let $u$ be the characteristic vector of the $m\times n$ rectangle (it is, in fact, the vector of size $t$ consisting only of $1$'s).

The conclusion we want to reach is that there is a vector $x$ of size $k$ with rational non-negative entries s.t. $Ax=u$, while the hypothesis tells us that for every real vector $y$ of size $t$ s.t. $y^tA\le 0$, we have $y^tu\le 0$. By the [url=http://planetmath.org/encyclopedia/FarkasTheorem.html]Farkas-Minkowski Lemma[/url], we conclude that there must indeed exist a [i]real[/i] vector $x\ge 0$ s.t. $Ax=u$. We'd be done now if we could prove the following:

Given a rational $t\times k$ matrix $A=(a_{ij})$ and a rational vector $u=(u_i)_{i=1}^t$, if the system $\{\begin{array}{c}Ax=u\\x\ge 0\end{array}$ has a real solution tilde $x=(\tilde x_j)_{j=1}^k$, then it has a rational solution $x=(x_j)_{j=1}^k$

Given a real solution $\tilde x$, we’ll only modify the positive coordinates of $\tilde x$ in order to get a rational solution $x$, so, since we’re only concerned with the columns of $A$ which get positive coordinates from $\tilde x$, we may, for our purpose, assume that all the coordinates of $\tilde x$ are positive, i.e. $\tilde x>0$. We can also eliminate rows from the linear system $\sum_{j=1}^ka_{ij}x_j=a_i,\ i=\overline{1,t}$ s.t. the remaining rows form a basis for the row space of the matrix $A$ without affecting the result, because we know that the system is compatible. Instead of introducing other indices, we assume that the rows of $A$ are linearly independent (in particular, $t\le k$). By a renumbering of the columns of $A$, we can assume that the columns $A^1,\ldots,A^t$ form a basis for the column space of $A$. If we give arbitrary values to $x_{t+1},\ldots,x_k$, the values of $x_1,\ldots,x_t$ will be uniquely determined by Cramer’s rule. Now, if $x_{t+1},\ldots,x_k$ are close enough to $\tilde x_{t+1},\ldots,\tilde x_k$ respectively, then $x_1,\ldots,x_t$ will be close to $\tilde x_1,\ldots,\tilde x_t$ respectively, so, in particular, they will be positive. We simply choose $x_{t+1},\ldots,x_k$ to be rational and very close to $\tilde x_{t+1},\ldots,\tilde x_k$, and we’ll have the desired rational solution $x=(x_j)_{j=1}^k$.

P.S.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=6056&sid=6e970c6ee903c068c18f10f93cb2ee69]This topic[/url] is basically concerned with the same problem.


Let x_1, x_2, ..., x_2004 be real numbers, satisfying -1 <= x_i <=1 for all i=1, 2, ..., 2004 and Sum (x_i)^3 = 0. Find the maximal value of Sum x_i.

Note that it is problem for the year 2004. Do hurry to solve it, because in 2007 it becomes trivial.

Namdung

well i dont know if this is correct
x_i = sin (y_i)
the given condition  \sum (x_i)^3 implies
1/3  \sum sin (3y_i) =  \sum sin (y_i) =  \sum x_i
that implies that the maximum possible value is 2004/3 = 668
and when is the maximum achieved? :?
is it correct :)

You were right with estimation Sum <= 668. But unfortunately, this value can not be achieved, so 668 is not the answer.

Namdung

This inequality is very interesting and difficult. Can you please post a solution, Namdung?

The idea is no new, I just discribe in brief

We will solve the problem: Let be given 2004 real number x_i from
[-1,1] with sum 0. Find the maximum of  F(x*) = Sum (x_i)^(1/3)

Put f(x) = x^(1/3). We can prove following lemmas

1. If -1 < x, y < 0 then f(x)+f(y) <= f(-1) + f(x+y+1)
2. If x_1, ..., x_k > 0 then Sum f(x_i)  <= kf(Sum x_i /k)
3. If x > 0, y < 0 and
        a) x+y < 0 then f(x) + f(y) <= f(-1) + f(x+y+1)
        b) x+y> 0 then  f(x) + f(y) <= f(0) + f(x+y)

From these lemmas, we can conclude that if x# = (x_1, x_2, ..., x_2004) is the point where f(x*) takes maximum then all negative x_i are equal to -1 and all positive x_i are equal to each other.

Let we have k numbers -1 and 2004-k number a then
     -k + (2004-k)a = 0 => a = k/(2004-k)
F(x#) = -k + (2004-k)(k/(2004-k))(1/3) = G(k)

An easy analysis show that G(k) (k positive integer) takes maximum when k=223.

Answer: -223+223^(1/3)*(1781)^(2/3) ~ 667.999 so Galois is very near with right answer. In 2007 Galois solution will work as the best.

Namdung

I could have tried for years and I still couldn't have solved this magnificient problem. It is extraordinary!

boy,was i close??!! :D  :D
i had been working on the problem since yesterday and had got 2 of the lemmas but the solution is really coool,Namdung


What is the time taken to cover half the distance S1 if time taken to cover 2*S1 is t.

It is $\frac{t}{4}$.However this doesn't belong in pre-olympiad and for other physic problems use the introudctory physics forum  :)


algebra
polynomial
function
algebra proposed

A sequence  $(P_{n})_{n=0}^{\infty}$ of polynomials  with complex coefficients is called a  [b] polynomial sequence[/b] if and only if
[b] degree[/b]$(P_{k})= k$  for all  $k\in \{0,1,...\}.$ [i] Find   all polynomial sequences  $(P_{n})_{n=0}^{\infty}$ having the property [/i] \[P_{n}\left(P_{k}(x) \right) = P_{k}\left(P_{n}(x)\right) = P_{nk}(x)\; \; ,\; \; \forall k,n \in \{0,1,...\}.\]

I don't know how difficult it is to prove the special case above, but all polynomials with $P(Q(x)) = Q(P(x))$ are classified (over any field of characteristic 0).  Up to linear changes of variable, there are no examples except the well-known ones.

Yes, the two families are :
1) $,X,X^2,X^3, \cdots$
2) The Tchebychev polynomials $T_n$.

I think this result is known as Julia-Ritt's theorem. This has been generalized by Jacobsthal.
I don't have a proof which was not long and quite boring.

Pierre.

Once I wrote number of problems about commuting polynomials,
and they also include this problem.
However, they does not include the Julia-Ritt's theorem,
because this theorem uses complex dynamics or something like that,
but my problems have elementary solutions.
May be this is another chance to invite people to think about them.
The link is :
http://www.mathlinks.ro/Forum/viewtopic.php?t=30555

P.S.: Of course all the problems follow from this Julia-Ritt's theorem,
because it characterizes completely any pair of commuting polynomials,
but on the other hand the problems can be solved by purely elementary methods.

[quote="fleeting_guest"][quote] ... Julia-Ritt's theorem. This has been generalized by Jacobsthal. [/quote]

What is the generalization (and was not Jacobsthal before Ritt, who proved this theorem and its version for rational functions)?[/quote]
I ask for exact reference regarding Julia-Ritt's theorem. Also for Jacobstahl result. There are accesible on-line ?
Thanking in advance,Alex


If a, b, and c are real numbers and $a\ne0$, then $ax^2+bx+c$ can be written as $(x-\frac{-b+\sqrt{b^2-4ac}}{2a})(x-\frac{-b-\sqrt{b^2-4ac}}{2a})=0$ Use this result to discuss the graphs of the possible solutions of the inequality $ax^2+bx+c\le0$.

[quote="236factorial"]If a, b, and c are real numbers and $ a\ne0 $, then $ ax^2+bx+c $ can be written as $ (x-\frac{-b+\sqrt{b^2-4ac}}{2a})(x-\frac{-b-\sqrt{b^2-4ac}}{2a})=0 $ Use this result to discuss the graphs of the possible solutions of the inequality $ ax^2+bx+c\le0 $.[/quote]
This is because $ (x-\frac{-b+\sqrt{b^2-4ac}}{2a})(x-\frac{-b-\sqrt{b^2-4ac}}{2a})\le0 $ and so one binomial factor is negative and the other is positive.

Solution set is either $ S=\{x\in\mathbb{R}|x\le\frac{-b+\sqrt{b^2-4ac}}{2a}\right\}\cup\{x\in\mathbb{R}|x\ge\frac{-b-\sqrt{b^2-4ac}}{2a}\right\} $

or $ S=\{x\in\mathbb{R}|x\ge\frac{-b+\sqrt{b^2-4ac}}{2a}\right\}\cup\{x\in\mathbb{R}|x\le\frac{-b-\sqrt{b^2-4ac}}{2a}\right\} $


limit
integration
trigonometry
calculus
calculus computations

Calculate
$\lim_{x\rightarrow 0}\frac{\int_{1}^{x^{2}}\frac{\sin t}{t}dt}{\int_{1}^{x^{2}}te^{\sqrt{t}}dt}$

Edit: Sorry, it's a misprint in my calculus book (but I don't see how we could have used integration by parts or the squeeze theorem here).

Just use integration by parts on $\frac{\sin t}{t}$ and $te^{\sqrt{t}}$

Sorry, are you sure that $\lim_{x\rightarrow 0}?$, I think that $\lim_{x\rightarrow 1}.$


AMC
USA(J)MO
USAMO
number theory unsolved
number theory

Denote with $S_n$ the sum of the first n prime numbers.Prove that there exist at least one perfect square in $[S_n, S_{n+1}]$

Let $p_1 = 2, p_2, ...$ be the primes. Denote by $f(m)$ the smallest perfect square less than or equal to $m$. Note that $S_n \geq 2 + 3 + 5 + 7 + ... + (2n-1) = n^2 + 1$. Therefore, $f(S_n) \geq n$.

Now, assume for the sake of contradiction that $l^2 < S_n < S_{n+1} < (l+1)^2$. Then, we have $2l-1 \geq S_{n+1} - S_n = p_{n+1}$. Observe that

$p_n + (p_n - 2) + (p_n - 4) + ... + (p_n - 2(n-1)) + 1 \geq p_n + p_{n-1} + ... + p_1 = S_n$

so that $p_n \geq \frac{S_n - 1}{n} + n-1 \geq \frac{l^2}{n} + n-1$. Moreover, $p_{n+1} \geq p_n + 2$. Combining what we have so far yields

$2l-1 \geq \frac{l^2}{n} + n + 1$, or $0 \geq l^2 - 2nl + n^2 + 2n \geq (l-n)^2 + 2n$, clearly a contradiction.

This is very similar to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=320842&highlight=#320842]USAMO 1994 Problem 1[/url], where we are not only restricted to prime numbers.

I think it appeared in an issue of Komal


At a diving competition, Alex and Brenda received the scores shown. Each diver's final score is determined by dropping the highest and lowest scores, calculating the sum of the remaining scores, and multiplying that sum by 0.6. What is the positive difference between the final score of Alex and the final score of Brenda? Express your answer as a decimal to the nearest tenth.
$ \begin{tabular}{c|c|c|c|c|c|c|c|}
\hline
Alex &9.0&9.0&8.5&10.0&8.0&9.5&9.5\\
\hline
Brenda &9.3&9.9&10.0&9.5&9.7&8.0&8.6\\
\hline
\end{tabular}$

Dropping Alex's highest and lowing scores, we have $ (9 \plus{} 9 \plus{} 8.5 \plus{} 9.5 \plus{} 9.5)(.6) \equal{} (45.5)(.6).$
Droppinmg Brenda's highest and lowest scores, we have $ (9.3 \plus{} 9.9 \plus{} 9.5 \plus{} 9.7 \plus{} 8.6)(.6) \equal{} (47)(.6)$.

So, we have $ (47)(.6) \minus{} (45.5)(.6) \equal{} .6(47 \minus{} 45.5) \equal{} .6(1.5) \equal{} \boxed{.9}$.

Alex's scores that count toward his final scores are 9.0,9.0,8.5,9.5,9.5.  Their sum is 43.5, and multiplying by 0.6 we get 27.3.

Brenda's scores that count toward his final scores are 9.3,9.9,9.5,9.7,8.6.  Their sum is 47, and multiplying by 0.6 we get 28.2.

Thus, their positive difference is 28.2-27.3=0.9


Show, for all positive integers $ n = 1,2, . . .,$ that 14 divides $ 3^{4n+2}+5^{2n+1}$.

Solution:

[hide]

Base case: n=0 (this ultimately will NOT violate the positive integer condition). That works.

Assume $ 7|(3^{4n+2}+5^{2n+1})$. Then $ 3^{4n+2}\equiv{a}\mod{14}$ and $ 5^{2n+1}\equiv{-a}\mod{14}$. Substituting $ n+1$ for $ n$ gives $ 3^{4n+6}+5^{2n+3}\equiv{81a-25a}\equiv{56}\equiv{0}\mod{14}$, and the induction is complete.[/hide]


(1 + sin x - cos x) / (1 + sin x + cos x) = tan (x/2)

$ 1\minus{}\cos x\equal{}2\sin ^ 2 \frac{x}{2}, 1\plus{}\cos x\equal{}2\cos ^ 2 \frac{x}{2}$ and $ \sin x\equal{}2\sin \frac{x}{2}\cos \frac{x}{2}$


analytic geometry
graphing lines
slope
trigonometry
function
geometry
geometric transformation

I'm not sure if this is the correct section, but this is the math I am currently in so here it goes. I'm trying to find an equation that would calculate a particle reflecting off of a surface of any slope. The input will be the particle's y and x velocity and the angle the surface is leaning relative to x-axis. The output will be the particle's new x and y velocity. I dunno why I'm having difficulty with this, but I can do the really simple ones such as a particle reflecting of of a surface perpendicular to the x-axis (90 Degrees) which the output would be -x and y. Or a surface parallel to the x-axis(180 or 0 Degrees), the output being x and -y. With surfaces that have odd slopes I would have to use trig functions, and I got answers that sounded right but they didn't apply to all angles of the surface. I need an equation that will work with negative or positive x and y velocities, all slopes (or within reason) of this surface that is reflecting the particle, and is efficient. I thought it would be easy, and I feel kind of bad since I bet it is easy.

All you need is to reflect the velocity vector around the surface. Suppose that $ \beta$ is the angle the surface makes with the horizontal axis. Now, the symmetry operator is given by the matrix $ \begin{bmatrix}\cos 2\beta&\sin 2\beta \\ \sin 2\beta& \minus{}\cos 2\beta\end{bmatrix}$, so
\[ x'\equal{}x\cos 2\beta\plus{}y\sin 2\beta\qquad y'\equal{}x\sin 2\beta\minus{}y\cos 2\beta\]

Thank you, but I figured it out an equation that is equal to the one you gave me. I did it by rotating the line so it is parallel to the x-axis, do the reflection by y becoming -y, and then bring the surface back to its original position. I wrote out the entire equation, canceled out and simplified, and ended up with something almost exactly like yours, but the big difference is that I defined theta as 180-Angle of Surface. But yours just uses the angle of the surface without that calculation, so I'm going to be using that equation in the format of mine. Thanks for introducing rotational matrices to me though.

[quote="Peacer"]Thank you, but I figured it out an equation that is equal to the one you gave me. [/quote]
So, why did you say that you
[quote]
got answers that sounded right but they didn't apply to all angles of the surface.
[/quote]
then? :? This equation works no matter what the velocities and the angle are....

You misunderstood me. I got the equation I found AFTER I posted.


algebra
polynomial
calculus
derivative
modular arithmetic
number theory
relatively prime

This problem is designed by Dr. Mohsen Jamali. A hard and beautiful problem that was proposed for Iran TST 2004.

Find all polynomials $p(x)\in\mathbb Z[x]$ that $p(n)|p(m)\Longrightarrow n|m$

Just a stupid observation: $p$ must be divisible by $X$, by taking arbitrary $n$ and $m=n+p(n)$.

I think it is not hard to finish the problem if someone is able to prove the following claim: for any nonconstant polynomial $f \in{\mathbb Z}[X]$ with $f(0) \neq 0$ and any positive integers $k,N$, there is a prime number $p>N$ such that $p^{k}|f(m)$ for some integer $m$.

If the claim is true, write $p=X^{k}q$, $X\nmid q$ and if $q$ is not constant, apply the lemma to deduce that there is some prime $p>|q(0)|$ such that $p^{k}|q(m)$ for some $m \in{\mathbb Z}$. It is clear that $p \nmid q(p)$, hence there is some $r \in{\mathbb Z}$ such that $r \equiv m(\bmod p^{k})$ and $r \equiv p(\bmod q(p))$, therefore $p^{k}|q(r)$ and $q(p)|q(r)$, so $p(p)=p^{k}q(p)|q(r)|p(r) \Rightarrow p|r \Rightarrow p|m$, so $p|q(0)$, a contradiction. Hence $q = a \in{\mathbb Z}$, so $p=aX^{k}$.

That's not difficult:obviously, one can suppose that the polynomial is irreducible. By Schur's theorem, there are infinitely many primes that divide some $f(n)$. Since $f$ and its derivative are relatively prime, for big prime $p$ which divides some $f(n)$, $p$ will not divide $f'(n)$. Then we can apply Hensel's lemma to conclude.

[i]To dezeta[/i]
Your approach is close to mine.

[i]To harazi :[/i]
It's not obvious for me that $f$ must be irreducible, (e.g. How do you get it for $f(x)=ax^{k}$?) maybe I'm idiot now :(
Problem is really difficult. 2 years a go I was the only one in the Summer Math Camp that solved this problem. Nobody solved the problem in last Math summer camp.

It is not hard. If $p(m)=0$ then $\forall n \ p(n)|p(m)\Longrightarrow n|m$ It give m=0, because $p(0)|p(a_{0})$ p(0)=0.
Let $p(x)=a_{k}x^{k}+...a_{n}x^{n}$. Let $|p(n)|$ minimal, when $n\not =0$. Let $d=gcd(a_{k},a_{k+1},...,a_{n}$ then $(\frac{a_{k}}{d},\frac{p(n)}{d})=1$. Let $m=n+x\frac{p(n)}{d}$ then $p(n)|p(m)$, but $n\not |m$ if k<n. Therefore $p(x)=ax^{n}$.
By this metod can be proved that:
If $p(x)\in Q[x]$ and $\forall k\in Z$ $p(k)\in Z$, deg(p(x))=n, then following statements are equavalent:
(1) $p(k)|p(m)$ if and only if $k|m$.
(2) $k|m\Longrightarrow p(k)|p(m)$.
(3) $p(k)|p(m)\Longrightarrow k|m$.
(4) exist a: $p(x)=ax^{n}$.
There are nice:
following statements are equavalent:
(5) (p(k),(p(m))=1 if and only if (k,m)=1.
(6) $(k,m)=1\Longrightarrow (p(k),p(m))=1$.
(7) $(p(k),p(m))=1\Longrightarrow (k,m)=1$.
(8) $p(x)=x^{n}$.

OK, I am convinced that the problem is not difficult. I'll send my solution.

This is my solution. I hope it does not have any typo.

Obviously $x|p(x)$. Assume that $p(x)=x^{k}Q(x)$ that $gcd(x,Q(x))=1$. We prove the problem by contradiction. Suppose that $deg\ Q(x)\geq1$.

[i]Lemma 1:[/i]
For each $k$ there are infinitely many primes $p$ that there exist $n\in\mathbb N$ such that $p^{k}|Q(n)$ (Name these prime numbers the set $S_{k}$)

[i]Lemma 2:[/i]
There is an infinite set $A\subset S_{k}$ that $gcd(Q(n),n)=1$ for each $n\in A$. Also $A\cap S_{k}$ is infinite.
[b]Proof:[/b] You see $gcd(Q(n),n)=gcd(Q(0),n)$ and there are infinitely many primes in $S_{k}$ that does not divide $Q(0)$. (Because $Q(0)\neq0$)

[i]Lemma 3:[/i]
There are $m,n\in A$ that $n^{k}|Q(m)$ and $Q(n)|Q(m)$ and $m,n\notin\{-1,0,1\}$.
[b]Proof:[/b] Suppose $p\in S_{k}\cap A$. There is $n\in A$ that $p^{k}|Q(n)$ (We know that $gcd(Q(p),p^{k})=1$). By Chinese Remainder Theorem, there is $m\in\mathbb N\backslash\{-1,0,1\}$ that \[m\equiv n\pmod{p^{k}}\]  \[m\equiv p\pmod{Q(p)}\] Thus $p^{k}|Q(m)$ and $Q(p)|Q(m)$ and we are done$.\square$

Now consider $m,n$ in last lemma. $gcd(Q(m),m)$ and $gcd(Q(n),n)=1$. Now $n^{k}|Q(m)$ and $Q(n)|Q(m)$. So $p(n)=n^{k}Q(n)|Q(m)|p(m)$. Thus $n|m$ and $n|gcd(Q(m),m)=1$ and this is contradiction, because $n\notin\{-1,0,1\}$.
Q.E.D.


Let a regular 2004-gon be in a unit circle. Consider a set of quadrilaterals Q, all vertices of which coincide with some of the vertices of the 2004-gon, and the lengths of the sides and diagonals of which are not 1. Let R be a subset of Q, consisting of quadrilaterals, which contain the center of the circle inside of them. Prove that the number of elements of R is exactly the half of the number of elements of Q.

Anybody have ideas? Is the problem clear?

Yeah, problem is clear :)
One remark: unit circle is a circle of diameter 1 in this problem ;)

Suppose $ABCD$ contains circumcenter. Construct points $A'$, $B'$, $C'$ and $D'$ as it is described on Figure 1.
Then it is easy to prove that exactly one of quadrilaterals $ABD'C'$, $BCA'D'$, $CDB'A'$ and $DAC'B'$ defines [i]convex[/i] quadrilateral, which doesn't contain circumcenter. In the same manner we can define reverse transformation of quadrilateral, which doesn't contain circumceter.
Thus we have bijection between $R$ and $Q\setminus R$, therefore $|R|=|Q|/2$.


Is there an easy way to determine the angles of a triangle from the 3 sides, other than finding the area using Heron's, then dropping perpindiculars, then the two smaller angles?

In fact, the Law of Cosines figures in the proof of Heron's.


number theory
prime numbers
number theory unsolved

In my study, I found an interesting thing as folloing:

(1) $ 13,17,19,23,29,31,37,41,43,47$ are the consecutive $ 10$ prime numbers, and they have the number $ 30$ as their symmetric center,

(2) $ 137,139,149,151,157,163,167,173,179,181,191,193$ are the consecutive $ 12$ prime numbers, and they have the number $ 165$ as their symmetric center,

and my computer said,

(3) $ 8021749,8021753,8021759,8021771,8021789,8021791,8021801,$$ 8021821,8021831,8021833,8021851,8021863,8021869,8021873$
  are the consecutive $ 14$ prime numbers, and they have the number $ 8021811$ as their symmetric center.

So I would like to know that for every integer $ n > 2$, does there exist the $ n$'s consecutive prime numbers that have the symmetric center.

And I will conjecture 'Yes'.

As there exist arbitrary long arithmetic progressions in the primes, the answer is yes.

The conjecture concerns [b]consecutive[/b] primes, and as such it's likely to be false.  The result for $ 30$ is a consequence of the fact that a small prime is prime if and only if it is coprime to $ 30$, but I think it's very unlikely that such patterns continue to hold (and there is very little known about [b]consecutive[/b] primes in arithmetic progression).

For consecutive ones, Schinzel's hypothesis does the job  :D

Hi ZetaX and t0rajir0u.

Thank you for your replyings.

I want to know how Schinze's hypothesis works on this conjecture.

On http://en.wikipedia.org/wiki/Schinzel's_hypothesis_H , this is one generalization of conjectures such as the twin prime conjecture.

The examples I showed above is only even numbers of consecutive primes. But there exist examples of odd numbers of primes (whose crnter is a prime), too.

My friend , Kouichi Nakagawa , showed two pages about this problem.

The even case is [url=http://www.research.att.com/~njas/sequences/A055381]here[/url].
The odd case is  [url=http://www.research.att.com/~njas/sequences/A055380]here[/url].

But there is no proof . Is thiis so hard to prove?


geometry
rectangle
parallelogram
trapezoid
3D geometry
prism
sphere

Well, my Specialized High School admissions test is next week, and I was wondering, what do YOU guys think are a MUST-KNOW for a test on a ninth grade level?

Is that geometry?

Area of square
Area of rectangle
Area of parallelogram, trapezoid, triangle
Volume of prisms, cones, cylinders, sphere

If it also contains algebra:

Stuff with arithmetic, geometric sequences and series
Stuff with polynomials

Those usually appear in my experience.

Also, you need Pythagorean Theorem, Quadratic Formula, factoring, and completing the square.

Systems of equations and functions are probably useful too.


inequalities
geometry
circumcircle
geometry proposed

Let D,E,F be points on BC,CA,AB of triangle ABC respectively
 R is a circumradius of ABC
prove that \[ \left( {DE \plus{} EF \plus{} FD} \right)\left( {\frac{1}{{AD}} \plus{} \frac{1}{{BE}} \plus{} \frac{1}{{CF}}} \right) \ge \frac{{AB \plus{} BC \plus{} CA}}{R}
\]


I'm just curious and considering applying.

Not sure, but I'm from New York and am considering applying.

We accept students from all over the country.  Many of the campers are from California just because California has a lot of students to begin with.  There are many students from other states who already registered too.  I don't have the exact ratio at hand, but it doesn't really matter.  Anyone who likes math will be welcome at the camp and will have a good time.

[quote="selfanalyst"]I'm from Ohio and considering.

Speaking of proportions, did any girls attend the camp last year?  I would not want to be the only one.[/quote]

You don't need to worry.  Many girls already registered this year.


Ring Theory
number theory unsolved
number theory

Find all $x,y,z \in \mathbb{N}$ satisfying equation: $x^{2}+2y^{2}=z^{2}$

Factoring over $\mathbb{Z}[ \sqrt{-2}]$,

$(x+y \sqrt{-2})(x-y \sqrt{-2}) = z^{2}$

The details are annoying, but we can show that if $gcd(x,y,z) = 1$, then $x+y \sqrt{-2}$ is a square; in other words,

$\exists a, b \in \mathbb{Z}: (a+b \sqrt{-2})^{2}= (a^{2}-2b^{2})+2ab \sqrt{-2}= x+y \sqrt{-2}$

So $x = a^{2}-2b^{2}, y = 2ab$ generates the primitive solutions (giving $z = a^{2}+2b^{2}$).  The non-primitive solutions are found by multiplying by a constant.

Using http://www.mathlinks.ro/Forum/viewtopic.php?t=87739 is probably easier and doesn't use factoring (which gets into trouble if the $2$ is replaced by something bigger).

Certainly there are problems if the $2$ is replaced by something bigger, but since $\mathbb{Z}[ \sqrt{-2}]$ is a UFD I think it is permissible in this case  :D


calculus
integration
function
trigonometry
real analysis
real analysis unsolved

I have attached a pdf of output from MAPLE attempting to evaluate an integral of an exponential function containing a complex irrational trig function as the exponent. MAPLE could not evaluate it when integrating with respect to the hangle phi. I then transformed the integral via varible substitutions to hopefully get it in a form that might be executable, but no luck there either.

Any suggestions?

This integral is a portion of a solution of a radiation wave emanating in the vicinity of source that is configured as a disc. :?:

Do you have any reason at all to think there is a closed-form answer?


[quote="probability1.01"][quote="MysticTerminator"]in the 2006 picture of the USAMO winners, half of them are just looking at each other

what the heck?[/quote]

Dude... why am I looking at YASHA B K??? BTW he is from Connecticut... I lost[/quote]

He moved to NC.


$ u_1 \equal{} \sqrt {\frac {3a^2}{b^2 \plus{} bc \plus{} c^2}}$ , $ u_2 \equal{} \sqrt {\frac {3b^2}{a^2 \plus{} ac \plus{} c^2}}$ , $ u_3 \equal{} \sqrt {\frac {3c^2}{a^2 \plus{} ab \plus{} b^2}}$

PART $ 1$: $ u_1u_2 \plus{} u_2u_3 \plus{} u_3u_1 \geq 3$

Proof:

$ u_1u_2 \plus{} u_2u_3 \plus{} u_3u_1 \geq 3$ $ \Longleftrightarrow$

$ ab\sqrt {a^2 \plus{} ab \plus{} b^2} \plus{} bc\sqrt {b^2 \plus{} bc \plus{} c^2} \plus{} ca\sqrt {c^2 \plus{} ca \plus{} a^2} \geq$

$ \sqrt {(a^2 \plus{} ab \plus{} b^2)(b^2 \plus{} bc \plus{} c^2)(c^2 \plus{} ca \plus{} a^2)}$ $ \Longleftrightarrow$

$ 2a\sqrt {(a^2 \plus{} ab \plus{} b^2)(a^2 \plus{} ac \plus{} c^2)}$ $ \plus{} 2b\sqrt {(b^2 \plus{} ba \plus{} a^2)(b^2 \plus{} bc \plus{} c^2)}$ $ \plus{} 2c\sqrt {(c^2 \plus{} ca \plus{} c^2)(c^2 \plus{} cb \plus{} b^2)} \geq$

$ 3abc \plus{} 2(a^2b \plus{} ab^2 \plus{} b^2c \plus{} bc^2 \plus{} c^2a \plus{} ca^2) \plus{} a^3 \plus{} b^3 \plus{} c^3$

LEMMA$ 1$: $ \sqrt {(a^2 \plus{} ab \plus{} b^2)(a^2 \plus{} ac \plus{} c^2)} \geq a^2 \plus{} bc \plus{} \frac {ac \plus{} ab}{2}$

By using this lemma and Schur's inequality the proof ends.

PART$ 2$: $ u_1 \plus{} u_2 \plus{} u_3 \plus{} (2\sqrt {3} \minus{} 3)u_1u_2u_3 \geq 2\sqrt {3}$

Proof:

We will use part $ 1$.

LEMMA$ 2$: $ x$ , $ y$ , $ z$ are nonnegative real numbers with $ xy \plus{} yz \plus{} xz \equal{} 1$

$ x \plus{} y \plus{} z \plus{} (6\sqrt {3} \minus{} 9)xyz \geq 2$

Proof of Lemma$ 2$:

$ z \equal{} \frac {1 \minus{} xy}{x \plus{} y}$ , $ m \equal{} x \plus{} y$ , $ n \equal{} xy$

Inequality is equivalent to;

$ (m \minus{} 1)^2 \geq (6\sqrt {3} \minus{} 9)n^2 \plus{} (10 \minus{} 6\sqrt {3})n$

If $ n < \frac {6\sqrt {3} \minus{} 10}{6\sqrt {3} \minus{} 9}$ then the inequality is obviously true.

If $ n \geq \frac {6\sqrt {3} \minus{} 10}{6\sqrt {3} \minus{} 9}$

$ m \geq 2\sqrt {n} \geq 2.[\sqrt {\frac {6\sqrt {3} \minus{} 10}{6\sqrt {3} \minus{} 9}}] \approx 1.0616 > 1$

$ \Longrightarrow (m \minus{} 1)^2 \geq (2\sqrt {n} \minus{} 1)^2$

$ (2\sqrt {n} \minus{} 1)^2 \geq (6\sqrt {3} \minus{} 9)n^2 \plus{} (10 \minus{} 6\sqrt {3})n$ $ \Longleftrightarrow$

$ (\sqrt {n} \minus{} 1)(\sqrt {3n} \minus{} 1)^2((2\sqrt {3} \minus{} 3)\sqrt {n} \plus{} 1) \leq 0$ which is obviously true since $ z \equal{} \frac {1 \minus{} n}{m}$ and $ 0 \leq n \leq 1$

By using LEMMA$ 2$ and part$ 1$ we can easily show that the part$ 2$.

By part $ 1$ and part $ 2$ the proof of the main inequality is done.  :wink:

[quote="quykhtn-qa1"]Let $ a,b,c$ be nonegative real numbers,no two of which are zero.Prove that if $ 0 \le k \le \sqrt {2\sqrt {3} \minus{} 3}$ then we have:
$ (1 \plus{} k\sqrt {\frac {3a^2}{b^2 \plus{} bc \plus{} c^2}})(1 \plus{} k\sqrt {\frac {3b^2}{c^2 \plus{} ca \plus{} a^2}})(1 \plus{} k\sqrt {\frac {3c^2}{a^2 \plus{} ab \plus{} b^2}}) \ge (1 \plus{} k)^3$[/quote]
See here: http://can-hang2007.blogspot.com/2010/01/inequality-53-v-q-b-can.html

P/s: I wrote the problem under my name because I found this result before you post it here. Sorry.


$ {Problema}$ a aparut in nr.4/2001 al revistei "Kvant" si are enuntul urmator:

"In triunghiul $ ABC$ notam cu I centrul cercului inscris si cu $ A_{1}$, $ B_{1}$ si $ C_{1}$ punctele de tangenta ale cercului inscris cu laturile [$ BC$], [$ AC$] si respectiv [$ AB$].
Notam cu {$ P$}$ \equal{}{AA_{1}}\cap{BB_{1}}$, {$ M$}$ \equal{} AC\cap{A_{1}C_{1}}$ si cu {$ N$}$ \equal{} BC\cap{B_{1}C_{1}}$. Sa se arate ca: $ IP\bot{MN}$". $ ({A. Zaslavsky)}$

$ {SOLUTIA}$(mea): $ A$ este polul dreptei $ B_{1}C_{1}$, iar $ A_{1}$ este polul dreptei $ BC$ fata de cercul inscris $ \Rightarrow$ dreapta $ AA_{1}$ este polara punctului $ N$. In mod analog se arata ca dreapta $ BB_{1}$ este polara punctului $ M$. $ \Rightarrow$ dreapta $ MN$ este polara punctului $ P$, asa ca $ IP\bot{MN}$. $ \blacksquare$

O alta solutie, diferita ce cea data aici(si mai lunga) a aparut in nr.2/2002 din "Kvant". Revistele "Kvant" din anii 1970-2003($ {in}$ $ {lb.rusa}$) pot fi gasile la una
dintre adresele:
http://kvant.mccme.ru/
http://www.eknigu.org/lib//M_Mathematics/MSch_School-level/Zhurnal_Kvant/

[quote="mihai miculita"][color=darkred]In triunghiul $ ABC$ notam cu I centrul cercului inscris si cu $ A_{1}$, $ B_{1}$ si $ C_{1}$ punctele de tangenta ale cercului inscris cu laturile [$ BC$], [$ AC$] si respectiv [$ AB$]. Notam cu $ P\in{AA_{1}}\cap{BB_{1}}$, $ M\in AC\cap{A_{1}C_{1}}$ si cu $ N\in BC\cap{B_{1}C_{1}}$. Sa se arate ca: $ IP\perp{MN}\ .$[/color][/quote]
[color=darkblue][b]Solutie fara polaritate.[/b] Aceasta problema este clasica si ilustreaza utilizarea polaritatii ([b]teorema Hire[/b]) in rezolvarea problemelor de geometrie. Voi utiliza numai [b]teorema lui Pitagora[/b] si o caracterizare des folosita pentru perpendicularitatea a doua drepte : $ AB\perp CD\Longleftrightarrow AC^{2}\minus{}AD^{2}\equal{} BC^{2}\minus{}BD^{2}\ .$ In concluzie, demontratia oferita poate fi inteleasa si de elevii din clasele a VII -a !

[b]Demonstratie.[/b]

$ \blacktriangleright$ $ BI\perp MA_{1}$ $ \Longleftrightarrow$ $ BM^{2}\minus{}BA_{1}^{2}\equal{} IM^{2}\minus{}IA_{1}^{2}$ $ \Longleftrightarrow$
$ BM^{2}\minus{}\left(IB^{2}\minus{}r^{2}\right) \equal{}\left(MB_{1}^{2}\plus{}r^{2}\right)\minus{}r^{2}$ $ \Longleftrightarrow$
$ MB^{2}\minus{}MB_{1}^{2}\equal{} IB^{2}\minus{}IB_{1}^{2}$ $ \Longleftrightarrow$ $ \boxed{\ MI\perp BB_{1}\ }\ \ (1)\ .$

$ \blacktriangleright$ $ AI\perp NB_{1}\Longleftrightarrow$ $ AN^{2}\minus{}AB_{1}^{2}\equal{} IN^{2}\minus{}IB_{1}^{2}$ $ \Longleftrightarrow$
$ AN^{2}\minus{}\left(IA^{2}\minus{}r^{2}\right) \equal{}\left(NA_{1}^{2}\plus{}r^{2}\right)\minus{}r^{2}$ $ \Longleftrightarrow$
$ NA^{2}\minus{}NA_{1}^{2}\equal{} IA^{2}\minus{}IA_{1}^{2}$ $ \Longleftrightarrow$ $ \boxed{\ NI\perp AA_{1}\ }\ \ (2)\ .$

$ \blacktriangleright$ $ IM^{2}\minus{}IN^{2}\equal{}\left(MB_{1}^{2}\plus{}r^{2}\right)\minus{}\left(NA_{1}^{2}\plus{}r^{2}\right)\implies$

$ \boxed{\ IM^{2}\minus{}IN^{2}\equal{} MB_{1}^{2}\minus{}NA_{1}^{2}\ }\ \ (3)\ .$

$ \blacktriangleright$ Din relatiile $ (1)$, $ (2)$ obtinem :

$ \left\{\begin{array}{c}BB_{1}\perp MI\Longleftrightarrow PB_{1}\perp MI\Longleftrightarrow PM^{2}\minus{}PI^{2}\equal{} B_{1}M^{2}\minus{}r^{2}\\ \ AA_{1}\perp NI\Longleftrightarrow PA_{1}\perp NI\Longleftrightarrow PN^{2}\minus{}PI^{2}\equal{} A_{1}N^{2}\minus{}r^{2}\end{array}\right\|$ $ \implies$

$ \boxed{\ PM^{2}\minus{}PN^{2}\equal{} MB_{1}^{2}\minus{}NA_{1}^{2}\ }\ \ (4)\ .$

Din relatiile $ (3)$, $ (4)$ obtinem $ IM^{2}\minus{}IN^{2}\equal{} PM^{2}\minus{}PN^{2}$, adica $ \boxed{\ IP\perp MN\ }\ .$

[b]Observatii.[/b] Aceasta problema se poate imediat traduce echivalent la $ I_{a}$ etc.
De asemenea, se poate usor generaliza. Rezultatele vor fi surprinzatoare. Incercati ![/color]


With A,B,C is angles of pointed triangle.Find min
$P=tg_{A}+tg_{B}+2tg_{\frac{C}{2}}$

We have $tgA+tgB \geq tg(\frac{A+B}{2})$.
Thus $P \geq \frac{2}{tg\frac{C}{2}}+2tg\frac{C}{2} \geq 4$.The equality holds for $C=\frac{\pi}{2}$,$A=B=\frac{\pi}{4}$
Therefore $P_{min}=4$.


limit
integration
real analysis
real analysis unsolved

Calculate the limit: $ \displaystyle \lim_{n\to\infty} \frac{1}{n} \sqrt[n]{(n\plus{}1)\cdot (n\plus{}2) \cdot \ldots \cdot (n\plus{}n)}$.

If I'm not mistaken;

\[ \equal{} e^{\lim_{n\rightarrow \plus{} \infty}\frac {1}{n}\sum_{k \equal{} 1}^{n}\ln(1 \plus{} \frac {k}{n})}\],

so the point is to integrate $ \int_1^2\ln(u)du$.

Let $ x_{n}\equal{}\frac{(n\plus{}1)\cdots(2n)}{n^{n}}.$

Since $ \frac{x_{n\plus{}1}}{x_{n}}\rightarrow \frac{4}{e}$, we get that $ \sqrt[n]{x_{n}}\rightarrow \frac{4}{e}$.

or by stirling $ \lim ... \equal{} \frac {1}{n} (\frac {(2n)!}{n!})^{\frac {1}{n}} \to \frac{4}{e}$


anyone play nukezone?

I just started playing this game....

I think it's amazing....

what I was think we can do is make our own AoPS clan...

I think that would be AWESOME!!!

[quote="Jhyon8149"]What exactly is Nukezone?  :?[/quote]

Yeah... What exactly is this "Nukezone". FPS? MMORPG?

Hah. I used to play starkingdoms which is a game just like this...

Nukezone ID: The Horde (#383145)

[quote]What exactly is Nukezone? Confused[/quote]

a game without graphic...

more like using strategies....

and attacking people

if you want to try it,I'll send you the invitation....

just give me your address...

i might as well try it... jzhsieh@gmail.com


integration
calculus
function
algebra
domain
absolute value
calculus computations

Please find the volume of the solid limited by (would be great if somebody show me the solid!):
$ z\equal{}0$
$ y\equal{}0$
$ x\equal{}0$
$ z\equal{}9\minus{}y^{2}$
$ x\plus{}y\equal{}3$

Sorry about my poor english!!

The vertical planes $ x \equal{} 0$, $ y \equal{} 0$, and $ x \plus{} y \equal{} 3$ are "walls": they form an infinite triangular prism. Two other surfaces are the roof ($ z \equal{} 9 \minus{} y^2$) and the floor ($ z \equal{} 0$). Therefore, $ V \equal{} \int_0^3\int_0^{3 \minus{} y}(9 \minus{} y^2)\,dx\,dy$.

Hey man.. thx a lot! But I'm not able to do a good representation of this solid by hands. I'm having problems doing the roof... :(
By the way.. is there another way evaluating the volume? I mean using triple integral changing $ dV\equal{}dzdydx$ to $ dV\equal{}dxdydz$ or something like that...

Yes, I would use a triple integral if I knew that you got that far in multivariable calculus. $ V \equal{} \iiint 1\,dV \equal{} \int_0^3\int_0^{3 \minus{} y}\int_0^{9 \minus{} y^2}1\,dz\,dx\,dy$, or in some other order. The innermost integral can be evaluated at once: it's $ 9 \minus{} y^2$, and we are back to the double integral.

[[b]fedja[/b]'s picture should have $ 9$ instead of $ 3$ on the $ z$-axis.]

Yes, I menatlly inserted the square root for some reason and was quite perplexed as to why my parabolae were so close to circles :blush:. I corrected.

thanks a lot!!!! But I will always be able to calculate the triple integral in any order? Is there any restriction about that?

The key theorem is called Fubini's theorem; it states conditions under which the iterated integrals are equal (and the multiple integral makes sense). One sufficient condition is that the function and the domain are bounded. More generally, it works if the function is nonnegative, or if the integral of the absolute value (in either order) is finite.
Of course, if the domain isn't a box, you have to be careful about how the limits change.

A counterexample, in a case where none of the above conditions apply:
$ f(x,y)\equal{}\begin{cases}\frac1{y^2}&x<y\\ \minus{}\frac1{x^2}&x\ge y\end{cases}$ on $ [0,1]^2$
We get $ \int_0^1\int_0^1 f(x,y)\,dx\,dy\equal{}\int_0^1\left(\int_0^y \frac1{y^2}\,dx\plus{}\int_y^1\frac{\minus{}1}{x^2}\,dx\right)\,dy\equal{}\int_0^1 \frac1y\minus{}\frac1y\plus{}1\,dy\equal{}1$
By a similar computation, $ \int_0^1\int_0^1 f(x,y)\,dy\,dx\equal{}\minus{}1$.

thanks a lot about this help guys... cya

Why if I try calculation the volume like that: $ V \equal{} \iiint 1\,dV \equal{} \int_0^3\int_0^{3 \minus{} y}\int_0^{9 \minus{} y^2}1\,dz\,dx\,dy \equal{} \frac {81}{4}$ and if I try $ V \equal{} \iiint 1\,dV \equal{} \int_0^3\int_0^{9 \minus{} y^{2}}\int_0^{3 \minus{} y}1\,dx\,dz\,dy \equal{} \frac {135}{4}$. Where is my terrible mistake?  :maybe:

Not keeping track of which limits correspond to which variable. The second integral should be $ \iiint 1\,dV \equal{} \int_0^3\int_0^{9 \minus{} y^{2}}\int_0^{3 \minus{} y}1\,dx\,dz\,dy$.

Warning: in general, changing the order of integration is not as simple as in this example. It is advisable to practice with double integrals first.

My mistake... i've already corrected!Look at that again for me please???

Both integrals are equal to $ \frac{135}{4}$.

[quote="mlok"]Both integrals are equal to $ \frac {135}{4}$.[/quote]
But I'm thinking about what you said: it's dangerous to change the order of integration in triple integral because it could be more difficult than was in this example. Is there a theorem or something that tell when can we integrate in the order $ dxdydz$ or some other order? Or when we can't integrate $ dxdydz$ but we can do  like $ dydzdx$?? It's Fubbini's theorem?

The danger mlok was referring to is not that the integrals might not be equal- it's that you might write down the wrong limits and miscalculate, as you just did.
In order to understand the new limits, understand the region. This is much easier to do with double integrals, as you can draw the picture. Try this exercise:

Write down a $ dx\,dy$ integral equal to $ \int_0^1\int_1^{e^x}1\,dy\,dx$


In a paper we write $1,2,3,...,2006$
Take 4 numbers $a_1,a_2,a_3,a_4$ then place them down with the order $a_4,a_3,a_2,a_1$. (For example : We take 1,2,2005 and 2006 then we have a new order $2006,2005,3,4,...,2003,2004,2,1$)
Prove that we can't get the order $2006,2005,2004,...,3,2,1$

Your opeartion is a composition of two transpositions. So it doesn't change the parity of number of inversions in the sequence. But initial number of inversions is 0 and final number is $\frac{2006\cdot 2005}{2}=odd$.


group theory
abstract algebra
function
superior algebra
superior algebra unsolved

Using the usual test for being a subgroup, give proofs of the following:

1) {${z\in \mathcal{C}^{x}: \mid z \mid=1 }$} and {$e^{(}^{1}^{+}^{i}^{)}^{t}: t\in \mathds{R}$} are subgroups of {$\mathcal{C},$$.$) (where $\mathcal{C}^{x}= \mathcal{C}$\{0})

For the first function I assumed its the boundary of the circle with between (-1,1) and (-i,i) , and i get how identity, closure and self inverses are included.

I have no idea what the second function is and how to compute it .

Would appreciate help

Thnx

[quote="ueamath"]Using the usual test for being a subgroup, give proofs of the following:

1) {${z\in \mathcal{C}^{x}: \mid z \mid=1 }$} and {$e^{(}^{1}^{+}^{i}^{)}^{t}: t\in \mathds{R}$} are subgroups of {$\mathcal{C},$$.$) (where $\mathcal{C}^{x}= \mathcal{C}$\{0})
[/quote]

This happens to be an imoprant group called "1-torus" and written as $\mathbb{T}$.

But anyways, lets show it is a subgroup of $<\mathbb{C}^{*},\cdot >$.

1)If $z_{1},z_{2}\in \mathbb{T}$ then $|z_{1}\cdot z_{2}|=|z_{1}|\cdot |z_{2}| = 1 \cdot 1 =1$ so $z_{1}\cdot z_{2}\in \mathbb{T}$.

2)The element $z=1+0i$ is the multiplicative identity.

3)If $z \in \mathbb{T}$ then $|1/z| = 1/|z| = 1$ so $1/z\in \mathbb{T}$ and $z \cdot 1/z = 1/z \cdot z = 1$. Thats you inverse.

4)The associative property is certainly true cuz its true for the larger set $\mathbb{C}^{*}$

Thanks a lot, I understood this now, so i attempted the second function from the question and would appreciate if sum1 confirmed it .

[quote="ueamath"]Using the usual test for being a subgroup, give proofs of the following:

 and {$e^{(}^{1}^{+}^{i}^{)}^{t}: t\in \mathds{R}$} are subgroups of {$\mathcal{C},$$.$) (where $\mathcal{C}^{x}= \mathcal{C}$\{0}) [\quote]

For closure of multiplication, I did $e^{(}^{1}^{+}^{i}^{)}^{t_{^}1}$ $e^{(}^{1}^{+}^{i}^{)}^{t_{^}2}$=$e^{(}^{1}^{+}^{i}^{)}^{t_{^}1}+{t_{^}2}$ $t_{1}+t_{2}\in \mathbb{R}$

For existence of inverse:
$\frac{1}{e^{(}^{1}^{+}^{i}^{)}^{t_{^}1}}$ = $e^{(}^{1}^{+}^{i}^{)}^{-}{t_{^}1}$

Then I let $T_{1}$= -$t_{1}$ and $T_{1}\in \mathbb{R}$

As for identity :

A function times its inverse gives the identity element..

I think its right but I have a gut feeling its not that easy .. Could anyone confirm?

Thank

It is that easy; your second subgroup is the image of $(\mathbb{R},+)$ under a homomorphism.
The first is as well, although the homomorphism has a nontrivial kernel in that case ($t\mapsto e^{it}$).


limit
trigonometry
algebra unsolved
algebra

Find
$ \lim_{n\to \infty}(3\cos n\alpha\minus{}\sin n\alpha)$
with $ \tan\alpha\equal{}\frac{1}{2}$

Denote by $ u_n \equal{} 3 \cos (n\alpha) \minus{} \sin (n \alpha)$. Then, we could derive the recursive relation: $ u_{n} \equal{} 2 \cos\alpha u_{n\minus{}1} \minus{} u_{n\minus{}2}$.

It should be clear that the sequence does not converge in $ \mathbb{R}$. (Why?)

Oh sorry, I was wrong.

If we denote $ \lim u_n \equal{} L$ then (due to well known limit theorems) we must have $ L \equal{} 3 \cos \alpha L \minus{} L$. Then $ (2 \minus{} 3 \cos \alpha) L \equal{} 0 \iff L \equal{} 0$ (because $ \tan \alpha \equal{} \frac 12 \Rightarrow \cos \alpha \equal{} \pm \frac 2 {\sqrt {5}}$). If the sequence has limit then its limit is $ 0$. I 'm just not so sure if this sequence converges to 0.

here isthe solution :
we denote  $ cos\beta \equal{}\frac{3}{\sqrt10}$ and $ sin\beta\equal{}\frac{1}{\sqrt10}$
then we will get : $ u_n\equal{}\sqrt10 cos(\beta\plus{}n\alpha)$ this sequence clearly have no limit


Let the function $f(x)$ such that
$f(x)=\frac{sinx}{x}$ when $0<x\leq 1$
$f(x)=1$ when $x=0$.
Prove that:
$\frac{17}{18}<\int\limits_{0}^{1}<\frac{1703}{1800}$

[quote="phamduyhiep"]Let $f(x)$ be the function difined
$f(x)=\{ \begin{array}{l} sinx/x \ \text{when} \ 0<x\leq 1 \\ 1 \ \text{when} \ x=0 \end{array} $.
Prove that:
$\frac{17}{18}<\int\limits_{0}^{1}f(x) dx<\frac{1703}{1800}$[/quote]
---
R U Vietnamese?


This was #13 but surprisingly easy, ill post my solution later:

Find all 3 digit numbers that equal the average of all permutations of their digits. ie 111 = (111+111+111+111+111+111) / 6

Yep or 7a = 3b + 4c from which the solutions are readily found.


howdy,
looking for [b]interactive/engaging websites[/b] that you know about, not just ones that tell you the formula but something like: [url=http://www.mathwarehouse.com]interactive math site[/url] . In particular, check out [url=http://www.mathwarehouse.com/triangle]triangle activity[/url]?

Victor

I see.
But I don't have that much free time till April 9.


Daca $a,b\in\mathbb{R}$ notam $f_{a,b}: \mathbb{R}\rightarrow\mathbb{R}$ ,
$f_{a,b}(x)=ax+b$ si $G=\{f_{a,b}| a\in\mathbb{R}^{*}, b\in\mathbb{R}\}$

1. Sa se arate ca $(G,\circ)$ grup.
2. Demonstrati ca $N=\{f_{a,b}| b\in\mathbb{R}\}$ este subgrup normal al lui G.
3. Enuntati teorema intai de izomorfism pt grupuri.
4. Sa se arate ca  $G/N\approx(\mathbb{R}^{*},\cdot)$.
                 G/N este multimea cat sau factor.

Ma intereseaza doar 4., adica partea cu izomorfismul daca poate cineva
sa-mi arate, cu multumiri anticipate.

[quote="Paul V."]2. Demonstrati ca $N=\{f_{a,b}| b\in\mathbb{R}\}$ este subgrup normal al lui G.[/quote]
Afirmația asta e adevărată numai pentru $a=1$ (pentru $a \neq 1$ nu se poate, deoarece $f_{1,0}$, elementul neutru al lui $\left( G, \circ \right)$, nu se află-n $N$).

Morfismul căutat e $g : G \to \mathbb R^\ast, \, g \left( f_{a,b}\right) = a$. Avem $\ker g = N$ (deci e subgrup normal) și $G \slash \ker g \cong \text{Im}\, g = \mathbb R^\ast$.


I'm practicing on the old AHSMEs, but I can't find the rules anywhere. Can someone tell me
1) Time limit
2) Aids allowed
3) Scoring system
4) whatever else

I have a copy from my cousin's 1995 AHSME, give me some time to type it up.

1. DO NOT OPEN THIS BOOKLET UNTIL TOLD TO DO SO BY YOUR PROCTOR.

2. This is a thirty-question multiple choice test. Each question is followed by answers marked A, B, C, D, and E. Only one of these is correct.

3. SCORING RULES: You will receive 5 points for each correct answer, 2 points for each problem left unanswered, and 0 points for each incorrect answer.

4. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and calculators that are accepted for use on the SAT. No problems on the test will [i]require[/i] the use of a calculator.

5. Figures are not necessarily drawn to scale.

6. The answers to the problems are to be encoded on the AHSME ANSWER FORM. Check the blackened circles for accuracy and erase errors and stray marks completely. Only answers properly encoded on the answer sheet will be graded.

7. When your proctor gives the signal, begin working the problems. You will have [b]90 MINUTES [/b]working time for the test.

The calculators were first allowed on the 1994 AHMSE. The very first calculator-allowed AHSME, in 1994, was also an unusually easy one, with unusually high scores, leading some of the teachers I know to asssume that calculators just made the whole thing a whole lot easier. Of course, it wasn't really that - it's just that the test writers somehow came up with easier questions. By the late 90's, they were plenty tough.

Way back in my day (1969-1971), there were more than 30 questions on the AHSME and the questions came in several groups with different point values per question in each group. Blank answered earned zero points, and wrong answers earned a score of negative (1/4) times the question value. Negative total scores were possible and I knew students who got them.


When it is in progress, you say it when you have it.  What is it?

You may ask Yes/No questions.

I agree with fraenkel, but just in case if it's not right..

[hide]an idea?[/hide]

Neither of those are right.  Why aren't you asking yes/no questions about it?

The "when it is in progress" part is a bit abstract, and you probably wouldn't say that in that exact context.  But if I said it in the context you'd normally use it in, it might give away the answer.

Is it physical?
Is it abstract?
Is it tangible?

It depends on the context.  In some cases you can answer yes; in others you can answer no.  However, I will focus on one specific case.

Is it physical? [b]Yes[/b]
Is it abstract? [b]No[/b]
Is it tangible? [b]Yes, sort of[/b]

Is it something we'd encounter every day?
Is it a solid?
Is it a liquid?
Is it a gas?

[quote="cf249"]I've got it!

[hide]It's bingo. When you're playing bingo, you say "bingo" if you have a bingo.[/hide][/quote]

VERY close!  That's not the answer I had in mind.  You interpreted "when it is in progress" correctly:
[hide]"When it is in progress" = "When you play it"[/hide]

is it a chess move (i was thinking checkmate, draw offer, or a check)?

if it's not a chess move, I have another idea:

[hide]is it UNO?[/hide]


Find an injection $ {R}^2$-> $ {R}$. Is there an injection from the set of all real sequences to $ {R}$?

[b]Moderator edit: This belongs into Calculus/Analysis, not into Combinatorics![/b]

The following pdf document will also help:

http://www.math.upenn.edu/~kennardl/math170/LectureNotes/Day15.pdf


rasfoind variantele de bac am intalnit urmatoarea problema
fie $a_{n}$ un sir cu termenii $\pm 1$, si $x_{n}$ sirul definit astfel :
$x_{n}=a_{0}+\frac{a_{1}}{1!}+\frac{a_{2}}{2!}+\ldots+\frac{a_{n}}{n!}$.
la ultimul subpunct se cerea sa se arate ca $\lim x_{n}$ este un numar irational.
m`am gandit ca trebuie sa se aplice subpunctele anterioare (pe care nu le mai tin minte - varianta este 11 pentru cei interesati) insa nu prea am gasit nicio cale directa
sunt curios sa vad o solutie "de clasa" la aceasta problema

[hide="solutia mea"]
ideea solutiei mele este una clasica, inspirata din aproximarile rationale.
presupun ca $L=u/v$, cu $u,v$ naturale nenule. Atunci, fie $x_{n}=p_{n}/n!$. voi evalua diferentele :
$\left|\frac{u}{v}-\frac{p_{n}}{n!}\right|\leq \sum_{k={n+1}}^{\infty}\frac{1}{k!}<\frac{1}{n!\cdot n}$.
apoi inmultesc cu $n!\cdot v$ :
$\left|n!\cdot u-vp_{n}\right| < \frac{v}{n}$.
cantitatea din modul este un numar intreg, iar pentru $n\geq v+1$, membrul drept este subunitar, deci $n!u=vp_{n}$, adica $L=x_{n}$, contradictie.
[/hide]

O aplicatie draguta: $e \neq p+\sqrt{q}$ cand $p,q \in{\mathbf Q}$.

Presupunem ca $e=p+\sqrt{q}$ de unde rezulta ca $e^{2}-2pe+p^{2}-q=0$ dar $e$ este transcendent.

nu cred ca asta era ideea postului lui adi :) totusi, demonstratia ca $e$ e transcendent e destul de complicata

Ca e convergent se demonstreaza destul de simplu.
$ \sigma ,\varphi$ doua functii pentru care $ a_{\varphi (i)} \equal{} \minus{} 1,a_{\sigma (i)} \equal{} 1 \Rightarrow x_n \equal{} \sum {\frac {{a_{\varphi (i)} }}{{i!}} \plus{} \sum {\frac {{a_{\sigma (j)} }}{{j!}} \equal{} a_n \plus{} b_n {\rm }} }$;Ele sunt doua siruri monotone si marginite deci convergente deci $ x_n$ e convergent;
Presupunem ca  $ x_n  \to \frac{p}{q},y_n  \equal{} x_n  \plus{} \frac{1}{n},y_n  \downarrow \frac{p}{q} \Rightarrow a_0  \plus{} \frac{{a_1 }}{{1!}} \plus{} ... \plus{} \frac{{a_n }}{{n!}} < \frac{p}{q} < a_0  \plus{} \frac{{a_1 }}{{1!}} \plus{} ... \plus{} \frac{{a_n }}{{n!}} \plus{} \frac{1}{n} \Rightarrow 0 < pn! \minus{} q(a_0 n! \plus{} ... \plus{} a_n ) < \frac{q}{{(n \minus{} 1)!}} < 1$,pentru n suficient de mare,contradictie.


AMC
AIME
probability
AMC 12 B
AMC 12 A
AIME I

All the best for all those in this forum who are giving the AMC 12 B.

Continue ur discussion of ur scores in this thread.

P.S. IS the AMC 12 B usually harder than 12 A.

No; usually one is slightly harder than the other, but it's unpredictable which one will be harder.

Also, please don't publicly discuss anything about the test until the AMC forum is unlocked.

I got my school's score sheet from A before taking B today. It said I got a 118.5 on AMC 12 A, 10.5 points lower than what I expected. Should I request rescoring? (I mean, is there any penalty besides $ \$$5.00?)

@befuddlers:
The 12B is usually about the same difficulty as the 12A. However, the averages for the 12B might be higher since less motivated (and thus possibly lower scoring) schools don't take the 12B. But that is just a theory.
Last year my scores were 133.5 on the 12A and 129.0 on the 12B. This year, I got a 133.5 on the 12A. I will not post my 12B score until the AMC releases the answers.

@timwu:
If you don't mind spending the 5 dollars, I guess that you should.
If there is a penalty besides the 5 dollars, then it would be the possibility your score goes down after they rescore it.

Um will they post the answers like today?

Or tomorrow?

Can we discuss the answers and our scores only after the official solutions are out?

Thanks, Santosh. I'm going to get the rescoring form tomorrow.

@befuddlers: well, the official solutions are already out for those whose teacher purchased the pamphlets and disseminated to them after the test (my teacher decided to hold until tomorrow). However, according to AoPS, we may not publicly discuss any aspect of the competition until the AMC forum is unlocked (which will happen some time tomorrow, probably afternoon).

The official solutions are never made publicly available on the AMC website.  Instead, they are sent to the contest managers.

Just in case this isn't clear already :  Please do NOT discuss the AMC B in public until the AMC forum is unlocked!

Also, Santhosh, if you look in the AMC archives, you'll see that in the six years (before this one) when there have been both A and B contests, the A contest has had a higher average score two times, and the B contest has had an average score four times.  So at the moment, there is nothing to suggest that either one usually has higher scores than the other. :wink:

Ugh even my teacher did not give us the solutions.

Gah i hav to wait till tomorrow to know my score :(

Good excuse for not sleeping well today :rotfl:

Recap of my reply in AMC forum: 129 (1-24 skipping #17 minus #11 and #24), barring bubbling errors and other unforeseeable circumstances (as what happened on my 12 A). Kinda disappointed because for the majority of the time I was not concentrating (until #19 when I started seeing a bunch of quick problems)...

I should never ask for score before a test again.

I got a 129, too.  I'm kinda sad cuz I missed #1...  I thought bagels and muffins were the same, so I got 5.  The instant I read the question, I was like...wait...something's wrong, but I ignored that feeling and just bubbled in 5.  Then there was #20, which I miscounted on.  I skipped #24 and #25, being the cowardly mortal I am (jk :P).  So yeah, I tied Tim, so I'm very proud of myself (though I know he would have done better).

Look everyone, I didn't use "phail" in this post!  Hurrah for me!

@Miles: I didn't think that there was a statistically significant difference. Thanks for confirming this.

AMC 12B 135: missed #24, left #16 and #25 blank, personal best, yet still somewhat disappointing.

I guess I'll just shoot for double digits on the AIME and USAMO.

BTW: Does anyone who isn't from Walton/Dickerson want to guess Claus's AMC 10B score?  :wink:

Actually, no they would not.

Took the 10 this time and got 138. This time Edward beat me and got 144, as did Abubakar...

Wait, did Claus get a 150?  What a pro...

But come on guys, why take 10B?  Take the 12B!

Wow, Claus gets a 150!? That's just pwning...

Nice job, Sitan. (Technically you beat me since I got 108 last year on 12B as a freshman)

Yeah, I should shoot for a double-digit score on AIME too (though I will be very pleased if I can get 8+).

Why are we assuming things from what Sitan, of all people, said? (No offense Sitan, just a joke)

What are you talking about timwu? You always aim for a 15 on the AIME! I mean, the probability that you can guess correctly on whatever you didn't do is extremely small, but that's besides the point. You still aim for a 15, all the time.

I think people only need an 11 on AIME to qualify, regardless of earlier AMC scoring.

Um, no, they need like 8 or 9, methinks, not 11. ;\

So what [b]did[/b] Claus get?

[hide="Problem 19"]
The solution given by the AMC committee was beautiful. But i did not think along those lines........
$ f(n)  \equal{} n^4 \minus{} 360n^2\plus{}400$. Maybe this method will provide for a generalization i am not sure though.
$ f(1) \equal{} 41$ is prime. Now $ f(n) \equal{} f(n)  \equal{} n^4 \minus{} 360n^2\plus{}400 \equal{} n^2(n^2\minus{}360) \plus{} 400$. Implies $ f(2)........ f(18)$are negative. Now from $ f(19)$ all terms are positive. Then we note that if $ f(n)$ is prime $ n$ is odd and congruent to 1 mod 4. If a prime is of the form 1 mod 4 it can be written as sum of two positive integer squares. But$ 400 \equal{} (20)^2$. This implies $ n^2(n^2\minus{}360)$ is a square. Implies $ (n^2\minus{}360)$ is a square. The only solution in positive integers is $ n\equal{}19$. This can be proven rigorously by writing $ n^2\minus{}360\equal{}y^2$ and noting than n and y are positive odd integers. :D. Not quick for an AMC but........ [/hide]


geometry
trigonometry
calculus
calculus computations

The circle with radius 1 shown in the figure touches the curve [b]y=| 2x |[/b] twice. Find the area of the region that lies between the curves.  [geogebra]a34d8e355ed2c38c8935d64f73c73d5a6ff9f733[/geogebra]

Answer $ 1 \minus{} \frac {\pi}{2} \plus{} \cos ^{ \minus{} 1}\frac {1}{\sqrt {5}}$ or $ 1 \minus{} \frac {\pi}{2} \plus{} \sin ^ { \minus{} 1} \frac {2}{\sqrt {5}}$

or $ 1 \minus{} \frac {\pi}{2} \plus{} \tan ^ { \minus{} 1} 2$.

Could you please be more detailed and present the whole solution of the problem?
Thanks.

[quote="APSStudent"]Hey,
[quote="great math"]
It does not require calculus application because virtually, we can use the areas of an arc and right angled triangle. At this stage, calculus is so powerful and time consuming to carry out.
[/quote]
I'd like to see your solution not involving calculus.

Thanks,

-[u][b][url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url][/b][/u][/quote]Me too. It is exactly what i'm looking for. The solution involving calculus is pretty long, but I post it here, because I didn't know its appropriate place.

Because of the curves are symmetric through y- axis so we could possibly denote that the circle (I) with radius 1 and the tangents are the straight line ( $ y \equal{} |2x|$). Let the right tangent touches the circle at A.
We have $ \frac {IA}{OA} \equal{} tan(IOA) \Rightarrow OA \equal{} ?$ (can be directly calculated from the property of the line)
Let the circle cuts y axis at M.
We have to calculate the area of sector MIA ( wikipedia has this formulae http://en.wikipedia.org/wiki/Circular_sector).
After that, we evaluate the area of the right angled triangle IAO which is not so difficult to figure out.
By subtracting 2 areas shown above, we get the desired result.
The work I have shown is  just the geometrical interpretation.


Is there a simple solution for expression complex numbers in polar form for the application of De Moirvre's theorem? For example, the problem [tex]\displaystyle\frac{(1+ i)^6}{(1-i\sqrt3)^4}[/tex].

Thanks, that helps alot. I didn't quite follow the division before, but now it's a little clearer.


geometry
invariant
geometric transformation
rotation
analytic geometry
vector
trigonometry

Let A', B', C' be three points in the interior of a triangle ABC such that:

(1) A, B' and A' are collinear, with B' between A and A'.
(2) B, C' and B' are collinear, with C' between B and B'.
(3) C, A' and C' are collinear, with A' between C and C'.
(4) the four triangles AA'C, CC'B, BB'A, A'B'C' all have equal areas

If AB=c, BC=a, CA=b show that
\[ AA' = \frac{1}{4}\sqrt {\frac{{\left( {5 + \sqrt 5 } \right)}}{2}\left( {2a^2  + b^2  - c^2 } \right) + \sqrt 5 a^2  + c^2 } \]

[quote="fleeting_guest"]The linear transformation of this configuration onto an equilateral triangle should produce a symmetrical (invariant under 120 degree rotation) configuration.  From this you can calculate the barycentric coordinates of A', B', C'.  Then $|AA'|^2$ is a simple calculation with vectors and the Cosine Law in triangles.[/quote]

Could you show your calculations Fleeting ?  :blush:

Thanks in advance
Leon

I think, fleeting_guest didn't perform calulations. It was just a plan.

[quote="Leon"][quote="fleeting_guest"]The linear transformation of this configuration onto an equilateral triangle should produce a symmetrical (invariant under 120 degree rotation) configuration.  From this you can calculate the barycentric coordinates of A', B', C'.  Then $|AA'|^2$ is a simple calculation with vectors and the Cosine Law in triangles.[/quote]

Could you show your calculations[/quote]

The first part is a lemma to prove, and the last part reduces to knowing the definition of
the inner product of 2 vectors. Assuming the lemma, the bary.coordinates of A',B',C'
are cyclic permutations of C' = (x,kx,1/4) for some x and k.  Lines through A have coordinates (x,y,z) with y:z  in a fixed ratio.  You can work out the rest easily.


I'm having trouble answering some conceptual question about friction.  Could someone please check my responses and help me with the ones that I got wrong/don't know?  Thanks in advance!

[b]Should a line fitted to a maximum static friction force versus normal force graph pass through the origin?[/b]

A line fitted to the maximum static friction force versus normal force graph should pass through the origin.  If there is no normal force, then the object must not be resting on anything, and thus, there is no static friction.

[b]Should a line fitted to a kinetic friction forces versus normal force graph pass through the origin?[/b]

A line fitted to the average kinetic friction forces versus normal force graph should pass through the origin.  If there is no normal force, then the object must not be resting on anything, and thus, there is no kinetic friction.

[b]Does the coefficient of kinetic friction depend on speed?[/b]

The coefficient of kinetic friction should not depend on speed. Kinetic friction depends on acceleration, and acceleration depends on the change in speed over a given change in time.

[b]Does the force of kinetic friction depend on the weight of a box?  Explain.[/b]

I don't know this one.  Could someone please explain?

[b]Does the coefficient of kinetic friction depend on the weight of a box?[/b]

I don't know this one.  How is it different from the previous question?

To the last two questions:

Force of kinetic friction = coefficient of kinetic friction x Normal force

When weight of the box is more its counteracting normal force is also more. Thus force of kinetic friction increases. Hence the force of kinetic friction depends on the weight of the box.

But the coefficient of kinetic friction is a property of the material of the box and is a fixed value for any given material. Hence it does not depend on the weight of the box.

1.It should indeed pass through the origin as maximum static friction is given by $ f\equal{}\mu N$ which is equivalent to $ y\equal{}mx$

2.It must pass through the origin.

3.It is nearly independent of velocity(for high speeds it is not).

4.Not necessarily true.What if the block is pushed vertically upwards along a wall?

5.NO

There is a difference between force of friction and coefficient of friction.

[quote]4.Not necessarily true.What if the block is pushed vertically upwards along a wall? [/quote]

Could you make this case clear by listing out the forces involved in the free body diagram. Here as you said, obviously friction does not depend on weight..but what is the normal reaction in this ideal case..

While pushing the block upwards,the force cannot be vertical but at some angle $ \theta$ with the vertical.So the forces acting are:-
1.$ F cos\theta$ upwards
2.Gravity+friction downwards
3.Normal force outwards and
4.$ F sin\theta$ inwards

Thanks..

So in the case of pushing straight up with a vertical force the friction will be zero.. ( ofcouse only in ideal cases )

You cannot push a block straight upwards as there is an unbalanced normal reaction outwards which will try to push it away from the wall.So there must be a force which will balance the normal reaction,no matter whether there is friction or not.