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[quote="Tare"]I saw it on IMAX, and was wondering where everyone else saw it.[/quote]
Was it worth the IMAX?

**** yeah! (Excuse my language)
The highway chase scene was so...powerful.
When you actually see something that big, you see "in" to the picture. (Look up "Rules of perspective" for more info)

The other parts were kind of bad, though. I think it's not worth it seeing on normal screen, but it is worth it on IMAX.

The thing is, I don't really like watching IMAX movies, especially when they make you think that you are 'moving', if you get what I mean... it makes me feel dizzy...  :oops:

[quote="Tare"]Well I think IMAX 3D movies are better than Spy Kids 3D...
That movie made me color-blind :)[/quote]

What?  You mean you actually saw that one?  No offense to the makers, but that sounded, like, really dorky.

P.S. If anyone saw that on their computer, they deserve to die.  The only computer versions are, aside from being obviously pirated and illegal, a pain to download, so anyone that saw them must have really fast internet, which makes me jealous.   :)

Oh, and you're calling me a dork...</conversation>

Anyways, with compression of file size, dropping quality, and using mpeg4 technology, it's not that bad.

FDoHSI: I [b]did not [/b] see it on the computer, nor do I plan on seeing it on the computer since I already saw it. I [b]do not[/b] support piracy, nor have I ever done piracy.

I've never seen either, nor have I seen a movie on IMAX for that matter.

Movies in IMAX are pretty good - well, it depends on the movie itself as well...

I saw "Ghosts of the Abyss" on a field trip, and it was...well, educational. The other movie I saw was "Roller Coaster 3-D" or somat and it was pretty simple, but fun to watch.
The irony: "Ghosts of the Abyss" is 3-D and "Roller Coaster 3-D" isn't.

So watching an IMAX movie is just like watching a normal movie, except with a much larger screen, right? Hmm, and is Matrix Reloaded on IMAX shown in 2d or 3d?

Yes and no. IMAX theater can extend all the way to about 60 degrees from the ground to -30. It is curbed and it's like an inside out television. On the other hand, there are IMAX theaters that "claim" to be IMAX, but it's just a normal theater with abnormally big screen. That's why I seperated them in the poll.
As to if Matrix Reloaded is 3-D or not, no :( It would be awesome if it was, though.

Yes, Matrix in 3D would be a fantastic experience...

Think about it, Trinity coming out of the screen (and running you over :) )
That french guy :) throwing knives at you
And the explosions would be neato!

Course, the coolest part will be...*deleted by Wachowski brothers* :)

Ever wonder how the important people in the making of the movie are brothers?

So are real IMAX screens a lot better than *fake IMAX* screens in terms of viewing? And what do you think would make for a better first experience: seeing Matrix Reloaded in 2D or another ok movie in 3D? Hmm...though I might get a headache watching the 3D, heh (It happened once when I was at Disney World in Florida watching a short clip of honey i shrunk the kids).

By the way, is a "15/70 film frame" considered the real IMAX screen or the fake rectangular one?

Is the screen shown on this video the real one you were referring to or the fake one?

http://nz01.terabyte.co.nz/jordansimax/DocumentLibrary/imax_design.mpeg

Yes, Matrix Reloaded in 2D, it depends on whether its projected onto a flat surface/wall or a dome, hard to say. (Usually it's more rounded vertically too, not only horizontally. I'd say no, since if it was projected onto a dome it would be rounded in all direction. It looks as though they made the wall bent horizontally.)

Yeah, on the video, it appears that the screen is bent only on the left and right sides (though the narrarating voice says "down, up, and to the sides"). How much cooler to view is a screen that is rounded on all 4 sides compared to one that is not rounded at all?

The laws of perseption states that it's much easier to think that your actually there in the movie if the screen was bent (a 3-D arc), or somat.
The best would be to enter a completely round room (like a ball) where somat like a mirrorball project the movie all around you. Kind of like the room Profesor Xaviar enters to search out where people are (I forgot its name)

[quote]The laws of perseption states that it's much easier to think that your actually there in the movie if the screen was bent (a 3-D arc), or somat. [/quote]

Yeah, I know that that is the intended effect, but does it really work?

Kinds of. Depends on the movie. If it's a movie like "Finding Nemo" then yes, but not with Matrix Reloaded.
Movies where your in water or flying in air would go along great with the "dome" IMAX, but other movies are better off on the rectangular, fake IMAX.


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Hey, I've been working old Nationals problems from 1994-2005.  I can't get a lot of the questions, and I need help with them.  So any time I come across a problem I can't solve, I'll post it here for some help.  It'll be kind of like a marathon, but I will post all the problems. :P  I'll start with 1997's unsolved problems.  Some are quite easy, I just couldn't think straight at the time.  Anyway...

[u]1997 - Sprint #9[/u]
A broken toaster toasts two slices of bread at once but toasts only one side, so the bread must be flipped over to toast the other side.  It takes 45 seconds to toast one side.  What is the least number of minutes required to toast five slices of bread on both sides?  Express your answer as a mixed number.

[u]1997 - Sprint #27[/u]
Point P is on $\overline{AB}$.  Point A has coordinates $(3,11)$ and point B has coordinates $(18,1)$.  The ratio $AP: PB = 2: 3$.  What is the sum of the coordinates of point P?

[u]1997 - Sprint #30[/u]
What integer $n$ has  the property that $5^{96}$ is greater than $n^{72}$ and $5^{96}$ is less than $(n+1)^{72}$?

[u]1997 - Target #7[/u]
Of 6000 apples harvested, every third apple was too small, every fourth apple was too green, and every tenth apple was bruised.  The remaining apples were perfect.  How many perfect apples were harvested.

[u]1997 - Team #10[/u]
Given a 3x7 rectangular array of dots, how many triangles can be formed whose vertices are dots in the array>

[quote="mysmartmouth"][u]1997 - Sprint #30[/u]
What integer $n$ has  the property that $5^{96}$ is greater than $n^{72}$ and $5^{96}$ is less than $(n+1)^{72}$?[/quote]

[hide]
Consider $n^{72}<5^{96}\Longleftrightarrow n^3<5^4=625\implies n_\max=8$. Also $5^4<9^3$ so we're good. $\boxed {8}$.
[/hide]

[quote]
[u]1997 - Target #7[/u]
Of 6000 apples harvested, every third apple was too small, every fourth apple was too green, and every tenth apple was bruised. The remaining apples were perfect. How many perfect apples were harvested.
[/quote]

[hide]
Principle of Inclusion-Exclusion (PIE) gives the # that aren't perfect:

#(3)+#(4)+#(10)-#(3,4)-#(3,10)-#(4,10)+#(3,4,10)

=2000+1500+600-500-200-300+100

=3200

$\Rightarrow 6000-3200=\boxed {2800}$...
[/hide]

Edit: fixed

No the answer is 2800 for Target Number 7.  Anyone else wanna try....I'm confuzzled.

Am I allowed to speak even if I am from Michigan?

I can explain all these questions then.  And just to be sure, what is the answer to the broken toaster one?

[quote="mysmartmouth"]
[u]1997 - Sprint #9[/u]
A broken toaster toasts two slices of bread at once but toasts only one side, so the bread must be flipped over to toast the other side.  It takes 45 seconds to toast one side.  What is the least number of minutes required to toast five slices of bread on both sides?  Express your answer as a mixed number.[/quote]

[hide]In total we need to toast 10 sides of bread. Since the toaster can toast 2 sides at once it will take 5 total rounds to toast all sides. At 3/4ths of a minute per round it will take a total of 5*3/4 = [b]3.75 minutes[/b] to toast all sides. (Note: oo toast all of the sides in 5 rounds will require some jungling, but it is possible).[/hide]

[quote="mysmartmouth"]
[u]1997 - Sprint #27[/u]
Point P is on $\overline{AB}$.  Point A has coordinates $(3,11)$ and point B has coordinates $(18,1)$.  The ratio $AP: PB = 2: 3$.  What is the sum of the coordinates of point P?[/quote]

[hide]Since the point P is 2/5ths of the way from A to B its x-coordinate will be 2/5ths of the way from A's x-coordinate to B's x-coordinate (and likewise for the y-coordinates). So the x-coordinate of P will be $P_x = 3 + \frac{2}{5}(15) = 9$ and $P_y = 11 + \frac{2}{5}(-10) = 7$. Therefore $P_x + P_y = 16$[/hide]

[quote="mysmartmouth"]
[u]1997 - Sprint #30[/u]
What integer $n$ has  the property that $5^{96}$ is greater than $n^{72}$ and $5^{96}$ is less than $(n+1)^{72}$?[/quote]

[hide]$n^{72} < 5^{96} < (n+1)^{72}$ implies that $n^3 < 5^4 < (n+1)^3$. Therefore we are looking two consecutive cubes that straddle 625. 8^3 = 512 and 9^3 = 729, so the answer is $n = 8$.[/hide]

[quote="mysmartmouth"]
[u]1997 - Target #7[/u]
Of 6000 apples harvested, every third apple was too small, every fourth apple was too green, and every tenth apple was bruised.  The remaining apples were perfect.  How many perfect apples were harvested.[/quote]

[hide]Use the Principle of Inclusion Exclusion. If you don't know what that is you can find a discussion of it in the Art of Problem Solving book.

$6000 - \Big[\frac{6000}{3} + \frac{6000}{4} + \frac{6000}{10}\Big] + \Big[\frac{6000}{3 \cdot 4} + \frac{6000}{3 \cdot 10} + \frac{6000}{4 \cdot 10}\Big] - \Big[\frac{6000}{3 \cdot 4 \cdot 10}\Big]$
$= 6000 - (2000 + 1500 + 600) + (500 + 200 + 150) - ( 50)$
$= 2700$[/hide]

[quote="mysmartmouth"]
[u]1997 - Team #10[/u]
Given a 3x7 rectangular array of dots, how many triangles can be formed whose vertices are dots in the array>[/quote]

[hide]For any choice of 3 dots we can form a triangle. There are $\binom{21}{3} = 1330$ such choices. But some of these triangles will be degenerate because they lie on a straight line. We can see that there are 7 possible horizontal lines and $3*\binom{7}{3} = 35$ vertical lines. I haven't counted the "diagonal" lines yet, but once you find them you can subtract them also from the total.[/hide]

[quote="mathgeniuse^ln(x)"]Am I allowed to speak even if I am from Michigan?[/quote]

Yeah, I'm from MD, but I'm sure Mysmartmouth doesn't mind. :)

[quote="chess64"][quote="gauss202"][quote="mysmartmouth"]
[u]1997 - Target #7[/u]
Of 6000 apples harvested, every third apple was too small, every fourth apple was too green, and every tenth apple was bruised.  The remaining apples were perfect.  How many perfect apples were harvested.[/quote]

[hide]Use the Principle of Inclusion Exclusion. If you don't know what that is you can find a discussion of it in the Art of Problem Solving book.

$6000 - \Big[\frac{6000}{3} + \frac{6000}{4} + \frac{6000}{10}\Big] + \Big[\frac{6000}{3 \cdot 4} + \frac{6000}{3 \cdot 10} + \frac{6000}{4 \cdot 10}\Big] - \Big[\frac{6000}{3 \cdot 4 \cdot 10}\Big]$
$= 6000 - (2000 + 1500 + 600) + (500 + 200 + 150) - ( 50)$
$= 2700$[/hide][/quote]

$\gcd(4,10)\ne 1$[/quote]

Yes, Adeel (chess64) is right.  The correct answer is 2800 (not 2700), though I don't know how to get it...

I fixed it:

[quote="chess64"][quote]
[u]1997 - Target #7[/u]
Of 6000 apples harvested, every third apple was too small, every fourth apple was too green, and every tenth apple was bruised. The remaining apples were perfect. How many perfect apples were harvested.
[/quote]

[hide]
Principle of Inclusion-Exclusion (PIE) gives the # that aren't perfect:

#(3)+#(4)+#(10)-#(3,4)-#(3,10)-#(4,10)+#(3,4,10)

=2000+1500+600-500-200-300+100

=3200

$\Rightarrow 6000-3200=\boxed {2800}$...
[/hide]

Edit: fixed[/quote]

You do know PIE, right?

Um, no.  I don't know the Principle of Inclusion Exclusion.  Anyone want to explain?

Do you have the AoPS book? If not, just [url=http://mathworld.wolfram.com/Inclusion-ExclusionPrinciple.html]MathWorld[/url] it.

EDIT: Actually that might be a bit confusing...

Try this: http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=principle++inclusion&t=50743

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22841

Or just search "inclusion-exclusion".

[quote]If not, just MathWorld it.

EDIT: Actually that might be a bit confusing... [/quote]

Er...yeah I rarely refer anyone to mathworld if they're trying to learn something new.  I usually only use mathworld when I want an authoritative definition or something of that sort.

I'd look in the AoPS books first.  You have the new counting book, right?  I'd imagine it's covered in there.

I would suggest Wikipedia for finding anything mathematical on the web (at least, anything except for extremely specific things). Oddly enough, Wikipedia seems to be a superset of mathworld. They even have an article on the Duke Math Meet. I guess this is getting kind of off-topic though.

Where is PIE located in the AoPS books?  Is it in Volume 1 or Volume 2?  What chapter?

I think it's AoPS V2 17.3...but I can't say for sure...that would require me walking all the way upstairs :D  (I looked at the TOC online, btw)

[quote="easyas3.14159..."]gauss202, could you explain how you found the number of vertical lines so easily?(by the way, 35 isn't a multiple of 3). ;) Also, it there a way to quickly count diagonals?Thanks![/quote]

for the diagonals, it shouldnt be too hard to count them manually as it is a team round question and so time is not a huge huge issue what with the problems being divided up amongst various team members.

To find the number of vertical lines, what Mr. Neal (gauss) did was consider each column separately. In each column, there are $7$ points. $3$ of these are used to determine a degenerate triangle. Because there are also three columns, we multiply $7\choose3$  by $3$ to get the total number of vertical lines.


What about the complexe solutions ?! :D
 :cool:


Prove that from a set of ten distinct two-digit numbers (in the decimal system), it is possible to select two disjoint subsets whose members have the same sum.

[hide]The maximum sum of a subset is $ 90+91+\cdots+99=945$.  Then since the minimum sum is 0 (the empty set) and you cannot have any one-digit sums, there are $ 945+1-9=937$ possible sums.

Also, we know that there are $ 2^{10}=1024$ possible subsets.  Therefore, by the Pigeonhole Principle, there exist two subsets that have the same sum.

Assume that these two subsets are not disjoint; then remove from each subset the elements that are in both subsets.  Then the two remaining subsets are disjoint and have the same sum, so we're done.[/hide]


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a(1),....,a(n),b(1),........,b(n)>0
prove the inequality sum min{a(i)a(j,b(i)b(j)}=<sum min{a(i)b(j),a(j)b(i)}

It already appeared on the forum some time ago. The solution is available at http://www.kalva.demon.co.uk

P.S. here's the link to the old topic :D

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=usamo+hardest&t=15292


Find all natural solutions to $ 3x^2\minus{}2y^2\equal{}1$.

[hide]take the resolvent $ u^2\minus{}6v^2\equal{}1$ which has solutions,
$ u_n\equal{} \frac{1}{2}((5\plus{}2\sqrt{6})^n\plus{}(5\minus{}2\sqrt{6})^n) , v_n\equal{} \frac{1}{2\sqrt{2}} ((5\plus{}2\sqrt{6})^n\minus{}(5\minus{}2\sqrt{6})^n)$
so our equation has solutions,
$ x_n\equal{} u_n\plus{}2v_n , y_n\equal{} u_n\plus{}3v_n$[/hide]

Note that you need to prove that these are all solutions.

I though that all the solutions were given by the resolvent. Does there exist a criterion how to prove, that this are the only solutions?


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Let $ a$, $ b$, $ c$ denote the sides of a triangle. Show that the quantity
\[ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\]
must lie between the limits $ 3/2$ and 2. Can equality hold at either limits?

The first is [url=http://www.mathlinks.ro/viewtopic.php?search_id=593079618&t=30201]Nesbitt's inequality[/url]. Equality is achieved for an equilateral triangle.

The second writes
\[ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\le 2\]
But we know
\[ \frac{a}{b\plus{}c}<\frac{2a}{a\plus{}b\plus{}c}\iff a<b\plus{}c\]
so summing cyclically we get the required.
$ 2$ is never achieved. (unless we allow a degenerated triangle with two vertices coinciding  :) )

The inequality is equivalent to
2(a+b+c)(1/(a+b)+1/(b+c)+1/(c+a))>=9 or
((a+b)+(b+c)+(c+a))(1/(a+b)+1/(b+c)+1/(c+a))>=9
which follows from AM-HM Inequality. (Or one can use Cauchy Schwartz inequality)
Again from triangle inequality it follows that
b+c>s,c+a>s,a+b>s
Thus a/(b+c)+b/(c+a)+c/(a+b)<a/s+b/s+c/s=2s/s=2
as desired.

[hide=great]$I=\[ \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\]$\\
$I+3=2s\cdot( \frac{1}{2s-a}+\frac{1}{2s-b}+\frac{1}{2s-c})$\\
from Titus lemma we get $I+3\ge \frac{9}{2}$\\
we get $\boxed{I \ge \frac{3}{2}}$\\
also from triangle inequality we get $a+b>c$$\implies ac+bc>c^2\implies 2ac+2bc>c^2+ac+bc$\\
gives $\frac{c}{a+b}<\frac{2c}{a+b+c}$\\
so adding all symmetrically we get\\
$\boxed{I<2}$\\
hence $\boxed{\frac{3}{2}\le I<2} \blacksquare$[/hide]

[hide = Nice. Thanks.]
we have a<b+c (1)
b<c+a (2)
c<a+b (3)
then from (1) we got a+b+c<2(b+c) => 1\a+b+c >1\2(b+c)
(2) we got b +a+ c<2(a+c) => 1\a+b+c >1\2(a+c)
(3) we got c+b+a<2(a+b) =>1\a+b+c>1\2(a+b)
then muliply respectively by 2a, 2b ,2c equation (1) ,(2) and (3)
2a\a+b+c>2a\2(b+c) (1)
2b\a+b+c >2b\2(a+c) (2) we can simplify by 2 in the right side of each inequality
2c\a+b+c >2c\2(a+b) (3)
now (1)+(2)+(3)
2(a+b+c)\a+b+c >a\b+c +b\a+c+ c\a+b
simplify by a+b+c and we get result
2> a\b+c +b\a+c + c\a+b
i think that we can have equality
[/hide]


can someone give some algebraic theorems that can be used in combinatorics??? :?:

Here are two nice links:  :)

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=methods&t=28601
look at post #8 of Grobber - there's link to superb article about algebraic methods

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=combinatorial+nullstellensatz&t=19496#
look at posts of Vess for example


:blush:

I feel so dumb, but I really have no idea!

EDIT: Sorry. Apparently you mean mop. Here: [url]http://en.wikipedia.org/wiki/Mop[/url]

My apologies.

dude, jb05 does not get his own category

also, MOP is somewhat awesome. Look it up on AoPS wiki or http://www.unl.edu/amc/

[quote="not_trig"][quote="fibointepi"]Do you mean the MO[b]S[/b]P?[/quote]

No, he does not mean the "mosp," which, according to some, is a cross between a moth and a wasp. MOSP only exists in the language of:
1) Official people
2) Stupid people
3) Official and stupid people
4) jb05

so yeah.

It's MOP.[/quote]

http://www.unl.edu/amc/a-activities/a6-mosp/mosp.shtml

Since they made the contest, I think its MOSP

but MOSP just sounds wierd

So MOP

[quote="not_trig"][quote="jb05"]I am apparently now an official person. That is pretty cool.[/quote]

Of course, you are jokingly implying that you aren't stupid. :wink:[/quote]

Wait really? Yay I learned something today.

[quote]No, he does not mean the "mosp," which, according to some, is a cross between a moth and a wasp. MOSP only exists in the language of:
1) Official people
2) Stupid people
3) Official and stupid people
4) jb05

so yeah.

It's MOP.
[/quote]

By the way. I'm a girl. :]
I know, the name's confusing.
I think it's in another language.
It was my nickname when I was younger.

[quote="not_trig"][quote="jb05"]I am apparently now an official person. That is pretty cool.[/quote]

Of course, you are jokingly implying that you aren't stupid. :wink: Anyway, I think you do get your own category... a MOPper that calls it MOSP? What?[/quote]

no


$ P(x)$ is a polynomial such that $ P(1) \equal{} 1$ and

$ \frac {P(2x)}{P(x \plus{} 1)} \equal{} 8 \minus{} \frac {56}{x \plus{} 7}$

for all real $ x$ where the expressions are well defined. Compute $ P(\minus{}1)$.

What's the source of this problem?  It's definitely been posted before.  Write $ P$ as a product of linear factors; there are a lot of restrictions on what those factors can be.

[hide]$ P(x)$ is a polynomial such that $ P(1) \equal{} 1$ and
$ \frac {P(2x)}{P(x \plus{} 1)} \equal{} 8 \minus{} \frac {56}{x \plus{} 7}$

Multiplying both sides by $ (x\plus{}7)P(x\plus{}1)$, we get $ (x\plus{}7)P(2x) \equal{} 8(x\plus{}7)P(x\plus{}1) \minus{} 56P(x\plus{}1) \equal{} 8xP(x\plus{}1)$

Letting $ x \equal{} 0$, we get $ 7P(0) \equal{} 0$. So $ x \equal{} 0$ is a root of $ P(x)$.

Letting $ x \equal{} \minus{}1$, we get $ 6P(\minus{}2) \equal{} 0$. So $ x \equal{} \minus{}2$ is also a root of $ P(x)$.

Letting $ x \equal{} \minus{}3$, get $ 4P(\minus{}6) \equal{} 0$. So $ x \equal{} \minus{}6$ is also a root of $ P(x)$.

[Notice that trying to continue by letting $ x \equal{} \minus{}7$ gives $ 0P(\minus{}14) \equal{} 0$, which isn't informative about $ P(\minus{}14)$.]

So we know that $ P(x)$ has roots at 0,-2, and -6. Furthermore, if the leading term of $ P(x)$ is $ a_nx^n$, then the leading term of $ P(2x)$ will be $ 2^na_nx^n$. So the leading term of the quotient of $ P(2x)$ and $ P(x)$ will be $ 2^n$.

Since this quotient is given to be 8, $ P(x)$ must be a cubic polynomial. So 0, -2, and -6 must be the only roots of $ P(x)$. Therefore, $ P(x) \equal{} ax(x\plus{}2)(x\plus{}6)$, for some real number a. Using $ P(1) \equal{} 1$, we find that $ a \equal{} \frac{1}{21}$. Therefore, $ P(\minus{}1) \equal{} \frac{1}{21}(\minus{}1)(1)(5) \equal{} \frac{\minus{}5}{21}$[/hide]


algebra
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Let $ a,b,c$ be positive integers. If $ 30|a\plus{}b\plus{}c$, prove that $ 30|a^5\plus{}b^5\plus{}c^5$.

Let $ x \equal{} a \plus{} b \plus{} c,y \equal{} ab \plus{} bc \plus{} ca,z \equal{} abc$,then
$ a^5 \plus{} b^5 \plus{} c^5 \equal{} x^5 \minus{} 5x^3y \plus{} 5xy^2 \plus{} 5x^2z \minus{} 5yz$.So it is enough to prove that $ 5yz$ is divisible by $ 30$,or
$ yz$ is divisible by $ 6$.Since $ a \plus{} b \plus{} c$ is divisible by $ 2$,we conclude that among $ a,b,c$ there is an integer divisible by $ 2$,so
$ yz$ is an even number.Now suppose that numbers $ a,b,c$ are not divisible by $ 3$.But $ 3|a \plus{} b \plus{} c$,so either $ a\equiv b\equiv c\equiv 1$,or $ a\equiv b\equiv c\equiv 2$,but then $ y$ is divisible by $ 3$,thus we are done.

Isn't it possible that, for example, $ x \equiv y \equiv 1 (\mod 3), z \equiv 2 (\mod 3)$?
Am I missing something?

P.S. How can I write correctly the formula above, i.e. with the left perenthesis not separated from the \mod?

Using Fermat's little theorem(that $ n^p \equiv n (\mod p)$ if $ p$ - prime), we have:
for every integer $ n$:
\[ n^5\equiv n (\mod 2) (evident) \\
n^5 \equal{} n^3\cdot n^2 \equiv n \cdot n^2 \equal{} n^3 \equiv n (\mod 3) \\
n^5 \equiv n(\mod 5)
\]
because 2,3 and 5 coprime numbers, we have that $ n^5 \equiv n (\mod 2\cdot 3 \cdot 5)$, and finally
\[ a^5 \plus{} b^5 \plus{} c^5 \equiv a \plus{} b \plus{} c (\mod 30)
\]

[quote="GioMott"]Isn't it possible that, for example, $ x \equiv y \equiv 1 (\mod 3), z \equiv 2 (\mod 3)$?[/quote]

No.  If $ z \equiv 2 \bmod 3$ then WLOG either $ a \equiv 1 \bmod 3, b \equiv 1 \bmod 3, c \equiv 2 \bmod 3$ or $ a \equiv 2 \bmod 3, b \equiv 2 \bmod 3, c \equiv 2 \bmod 3$.  In the first case $ y \equiv 2 \bmod 3$ and in the second case $ x, y \equiv 0 \bmod 3$.

In other words, $ (t \minus{} a)(t \minus{} b)(t \minus{} c) \equal{} t^3 \minus{} xt^2 \plus{} yt \minus{} z$ is a polynomial with roots in $ \mathbb{Z}/3 \mathbb{Z}$, so it cannot have coefficients that correspond to an irreducible polynomial.

Since $ \lambda(30)\equal{}4$ and $ 30$ is square-free, then $ a^5\equiv_{30}a$, $ b^5\equiv_{30}b$ and $ c^5\equiv_{30}c$, which means that $ a^5\plus{}b^5\plus{}c^5\equiv_{30}a\plus{}b\plus{}c$. So, if $ 30|a\plus{}b\plus{}c$, then $ 30|a^5\plus{}b^5\plus{}c^5$, QED.

[hide=My Tidy Proof]We know that \[6\ |\ a^3-a\]So we need to prove \[5\ |\  a^5-a\]With Fermat's Little Theorem \[a^{\phi(5)}\equiv 1 \pmod{5}\]\[a^{4}\equiv 1 \pmod{5}\]\[a^{5}-a\equiv 0 \pmod{5}\Longrightarrow 5\ |\ a^5-a\]\\ \[ 30\ |\  a^5-a\]\[30\ |\  b^5-b\]\[30\ |\  c^5-c\]Add them up\[30\ |\  \underbrace{(a^5+b^5+c^5)}_{\text{has to be divisible by 30}}-\underbrace{(a+b+c)}_{\text{divisible by 30}}\][/hide]


geometry
incenter
cyclic quadrilateral
geometry proposed

The diagonals $ AC$ and $ BD$ of a cyclic quadrilateral $ ABCD$ meet at $ X$. Let $ A',B',C',D'$ be the feet of the perpendiculars from $ X$ to $ AB,BC,CD,DA,$ respectively. Prove that: $ A'B'\plus{}C'D'\equal{}A'D'\plus{}B'C'$.

A'B'C'D' has a incircle with the incenter X


prove that 1/2<1/a+1/(a+1)+...+1/(a+a-1)<1 where a is a positive real integer greater than 1.

A hint:

[hide]How many terms are there in that sum?[/hide]

This problem is quite approachable for a beginning problem solver.

[quote="probability1.01"]For some reason I don't think this should be on "getting started"[/quote]

This thread will be promoted soon. It now has a new thread title so that people will know what it's about when it's moved.

[quote="probability1.01"]For some reason I don't think this should be on "getting started"[/quote]
for some reason, i was able to solve this with relative ease.

tokenadult wrote:probability1.01 wrote:For some reason I don't think this should be on "getting started"

This thread will be promoted soon. It now has a new thread title so that people will know what it's about when it's moved.

But the question is just asking to show that [hide]if you add n terms which are between 1/n and 1/(2n), you get bounds of 1/2 and 1, respectively.[/hide]

[quote="Silverfalcon"]1/2<1/a+1/(a+1)+...+1/(a+a-1)<1

Better to see:

[tex]\frac {1}{2} < {\frac {1}{a} + \frac {1}{(a+1)} + .... + \frac {1}{2a-1}} < 1  [/tex]

Right?[/quote]

Maybe bounding the series works...

Let 1/a+1/(a+1)+...+1/(2a-1) (when a is an integer greater than one) = S.

S has [i]a[/i] terms in it, and the least and greatest terms in it are 1/(2a-1) and 1/a respectively, so a/(2a-1) = a * 1/(2a-1) < S < a * 1/a = 1.

a/(2a-1) < S < 1

The limit of a/(2a-1) as a increases towards positive infinity is 1/2.

(a/a) / ((2a - 1)/a) = 1 / (2 - 1/a), as a increases, 1/a approaches 0.

Ta-da!
1/2 < S = 1/a+1/(a+1)+...+1/(2a-1) < 1
(1<a< +:inf:, where a is an integer)

[quote="zscool"]methinks you are making it a bit too complicated,

[tex]\displaystyle \frac{1}{2} = \frac{a}{2a} < \frac{a}{2a-1}[/tex]

which relies on the fact that 2a > 2a-1.[/quote]

Yeah that's what I did.  Your signature's a bit ironic though :-).

LoL!  I see the pun in the signature also.

Yeah, I was just trying to make my solution as rigorous as possible, I don't know why, I hate when that happens, but I still didn't see that insight when I did it.

How do you know there are a terms in the sum? Looks like ther should be a-1 terms.

How many integers are there from a to 2a-1 (where a is an integer)?

There are a terms.  You have to do (2a-1)-(a)+1, which is obviously a.  You can't just do (2a-1)-(a) because that doesn't count the first term.  Look at
1/1+1/2
There are 2-1+1 terms, not 2-1.


Euler
geometry
algebra
polynomial
analytic geometry
geometric transformation
reflection

Suppose that the cevians $ AP , BQ , CR$ meet at $ T$ . Prove that

                                   $ \frac{TP}{AP}$ $ \plus{}$ $ \frac{TQ}{AQ}$ $ \plus{}$ $ \frac{TR}{CR}$ $ \equal{} 1$

The Gergonne-Euler theorem (in one of its many forms), as seen in [url=http://www.cip.ifi.lmu.de/~grinberg/GergonneEuler.pdf]"An unexpected application of the Gergonne-Euler theorem"[/url] Theorem 2 (just because I'm too lazy to find a better link).

  darij

Can ne1 solve this using Ceva 's theorem??

This problem appear in Kiran Kedlaya's Geometry Unbound. I've just solved it using almost the same method that Darij, when $ P$ lies inside the triangle.

I didn't understand why we may assume without loss of generality that $ P$ lies inside the triangle. The same doubts ocurred when I saw the proof of Ceva's theorem. The proof still holds when the cevians meet outside the triangle (indeed, it's a statement about lines, not triangles), but why? Why can we be sure that the same proof holds in any configuration?

Maybe this would fit better in the "Geometry Theorems and Formulas" forum, but it is related with this problem, so I posted here. I'm sorry if I made a mistake.

@sahsiam
actually its not necessary that $ P$ should lie inside the circle since the question is general but since i think that the euler-gergonne theorem holds for any point in the plane of the triangle (not necessarily inside it).
in the proof of ceva's theorem even if you draw the figure for cevians meeting at an external point the proof will more or less be the same.I reckon you should give it a try.  :lol:  :D  :P

[quote="Sashsiam_2"]I didn't understand why we may assume without loss of generality that $ P$ lies inside the triangle. The same doubts ocurred when I saw the proof of Ceva's theorem. The proof still holds when the cevians meet outside the triangle (indeed, it's a statement about lines, not triangles), but why? Why can we be sure that the same proof holds in any configuration?[/quote]

In this case, you can rewrite the proof with [i]directed[/i] areas and [i]directed[/i] segment lengths, and then it will work for any point $ P$ anywhere in the plane.

Generally, often it happens that a geometric theorem can be written as a polynomial identity with respect to the (cartesian) coordinates of the points and lines involved. The Gergonne-Euler theorem is such a theorem (you can easily write it as an equation in the coordinates of the points $ A$, $ B$, $ C$, $ P$, and this equation will only consist of addition, subtraction, multiplication and division, so it is a polynomial identity). It is known that a polynomial identity holds if it holds when its variables lie in some open region of $ \mathbb{R}^n$; the condition "the point $ P$ lies inside the triangle $ ABC$" puts the coordinates of $ A$, $ B$, $ C$, $ P$ into such an open region, so proving the Gergonne-Euler theorem for the case of $ P$ lying inside of the triangle $ ABC$ already proves the Gergonne-Euler theorem in all cases.

I hope this was halfways understandable - I don't have too much time at hand. This deserves an explanation in a book or journal paper, but I don't remember having seen this anywhere. Maybe someone of the more active MathLinkers would write this up for Mathematical Reflections or Crux?

  darij


providing that $|b-\frac{a}{2}|< \frac{|a|}{2}$

prove that  $|b-a|<|a|$

You can use the triangle inequality $|x+y|\leq |x|+|y|.$
Rewrite $|b-a|$ as $|b-a|=\left|\left(b-\frac{a}{2}\right)+\left(-\frac{a}{2}\right) \right|.$


Eh...title says it all.  ___________ posted a problem that was on it, and I don't know if we are allowed to discuss it.

NOTE: Person was not me

No, seeing as many schools haven't taken it yet, including mine.

I don't think so...some schools didn't take it yet.

EDIT: Beaten

[quote="AIME15"]Hopefully nobody will go out and say "Hey!  _____ posted #x on the ____ test of the 2009 school round here!"[/quote]

Shouldn't it be deleted?  idk

i think someone should report it to the mods so no one gets an unfair advantage


algebra
polynomial
superior algebra
superior algebra unsolved

Let $ K \leq F$ an algebraic fields extension and $ f(x)\in F[X]$. Prove that there exists a nonzero polynomial $ g(x)\in F[X]$ such that $ f(x)g(x)\in K[X]$. Can you find a bound for the degree of $ f\cdot g$ if you know the $ [F: K]$ and the $ deg f$ ?

Alexandros

Set $ K\equal{}\mathbb R$ and $ F\equal{}\mathbb C$. Then $ F$ is algebraic over $ K$. Set $ f(x)\equal{}a\plus{}ibx$. Is there any such $ g(x)$ that $ f(x)g(x)\in\mathbb R[x]$?

Yes there is : $ g(x)\equal{}a\minus{}ibx \in \mathbb{C}[X]$. Then $ f(x)g(x)\equal{}a^2\plus{}b^2x^2 \in\mathbb{R}[X]$

Alexandros

Any suggestions for this problem?

Alexandros

If $ f$ is irreducible then write $ f(X)\equal{}\prod_{i\equal{}1}^r(X\minus{}z_i)$ where $ z_i$ are the roots of $ f$ in the algebraic closure of $ K$ (which contains $ F$).
Now $ f$ is the minimal polynomial of $ z_i$ over $ F$. Take $ h$ to be the minimal polynomial of $ z_i$ over $ K$. Then obiously $ f|h$. Denote $ g\equal{}\frac hf$. Now $ g(X)f(X)\in K[X]$.
If $ f$ is not irreducible decompose it in irreducible factors and apply the above method for each factor separately.

Thanks Kuba!
What about the upper bound of the degree of the polynomial $ h(x)$ in the first ($ f$ irreducible) and the second case ($ f$ not irreducible) if they give you the degree of $ f$ and the degree of the extension $ K\leq F$ ?

Alexandros


Gauss
LaTeX
modular arithmetic
number theory unsolved
number theory

Hello friends,

This problem is taken from iopc 2006 a programming event [b]but it is basically a mathematics problem.[/b]
The event IOPC 2006 is now over. So, now I hope u can answer me. Though there were many interesting problems in the event (and i will post em all, stay assured) this was the one I found interesting in particular. The problem bears the name Fermat and Gauss and here it is...

[color=blue]Problem
In the mansion of Mathematicus Supremus in heaven live all the great mathematicians of yesteryear's. However, even after losing their mortal self, these mathematicians have not stopped doing mathematics. They always come up with puzzles and riddles and baffle their fellow theoreticians with mind boggling questions. Gauss and Fermat happen to be among the most enthusiastic of these problem posers. Both are always trying to outsmart the other to establish their supremacy in the group.

The mathematicians of this mansion dine on a large round table. A number of chairs are arranged around the table. One night, Gauss and Fermat happened to reach for the dinner quite early. Losing no opportunity to show his supremacy, Fermat said: ``Hey Gauss, there is some time left. Why not play a game?''. Gauss, as always ready to outsmart Fermat, readily agreed. Fermat then said: ``Okay Gauss, the game will go like this. You will first sit on a chair. Then you will call out a number $a$ between $1$ to $n$, where $n$ is the number of chairs. I will then call out either clockwise or anticlockwise. If I call out clockwise, then you sit on the $a$-th chair in the clockwise sense from the chair you are presently sitting on, and if I call out anticlockwise you sit on the $a$-th chair in the anticlockwise sense. You will again choose a number, and I will again call out whether to move clockwise or anticlockwise, and so on. In this way, you have to sit on the maximum possible number of chairs''. Gauss replied: ``Hey Fermat, this is damn easy. I am sure that even any school kid will be able to play this game quite easily!''.[/color]

Well u need to tell the maximum number of chairs Gauss can sit on if the initial number of chairs equals n. Assume that both Fermat and Gauss play to the best of their intellectual capabilities....

[i]moderator edit: added the numbers/variables (the $\LaTeX$ parts)[/i]

Let $f(n)$ be the number we're looking for. It's easy to see that $f(1)=1, f(2)=2, f(3)=2, f(4)=4, f(5)=4, f(6)=4, f(7)=6, f(8)=8$. Some things can be guessed:

1. $f(2^c)=2^c$. Proof: we'll show that no matter where we are, we can eventually sit on any particular chair, which is clearly enough. Let the chairs be $0, 1, \ldots, 2^c-1 \pmod{2^c}$, where $0$ is the chair we want to get to. Suppose we're on a chair $k$ such that $2^x \| k$. Then Gauss says $2^x$, which will land him on a chair divisible by $2^{x+1}$. Eventually he will get to the chair divisible by $2^c$, that is, $0$. This is enough.

2. $f(p)=p-1$ for $p>2$. This is part of a more general statement, but we'll start with primes. First, obviously $f(p)<p$ since Fermat can always keep Gauss away from a particular seat. Now suppose that $f(p)=p-r$ with $r \geq 2$. Then take the case where Gauss has sit on $p-r$ chairs, and he will never reach any of the other $r$. Let the chairs be $0, 1, \ldots, p-1$ and the unreachables be $a_1, \ldots, a_r$ (these are residues $\pmod{p}$). Let $i(2)$ be the inverse of $2$. Note that if Gauss can get to $i(2)(a_i+a_j)$, then calling out $i(2)(a_i-a_j)$ he will reach either $a_i$ or $a_j$. Then $i(2)(a_i+a_j)$ mustn't be reached by Gauss either. Let $b_1, \ldots, b_k$ be the set of chairs that are either unreachable or which allow a finite number of moves that takes Gauss to an unreachable chair. This set must be closed under the operation $a*b=i(2)(a+b)$.

We will examine this claim now. Note that if $i(2)(b_i+b_j)=b_i$ then $b_i=b_j$, a contradiction, since the $b$s are distinct. Then $b_1*b_2, \ldots, b_1*b_k$ must be a permutation of $b_2, \ldots, b_k$. Adding and multiplying by $2$ we get $(k-1)b_1+b_2+\ldots+b_k=2(b_2+\ldots+b_k)$, so $kb_1=b_1+\ldots+b_k=S$. Similarly, $kb_i=S$ for all $i$. But then the $b_i$ are equal, a contradiction, unless $k=1$.

3. $f(n)=n-n/p$, where $n$ is odd and composite, and $p$ is the smallest prime divisor of $n$. To prove this, we use the same arguments as above. Indeed, as above, we can see that if $a_1, \ldots, a_r$ are the unreachables and $b_1, \ldots, b_k$ are the numbers that would lead to unreachables, then $b_1, \ldots, b_k$ must be closed under the operation $*$, so $kb_1=kb_2=\ldots=kb_k=S \pmod{n}$. However, we can't divide by $k$, since $(k, n)$ may not be $1$. Let $(k, n)=d, k=dp, n=dq$. Then the numbers $dpb_i$ are congruent $\pmod{dq}$, which happens iff the $b_i$ are congruent $\pmod{q}$. Since the $b_i$ are distinct $\pmod{dq}$ but all equal $\pmod{q}$, they must be pairwise distinct $\pmod{d}$. But there are $dp$ of them, so $p=1$ and the $b_i$ are a permutation of $0, 1, \ldots, d-1 \pmod{d}$. In particular $r \leq k=d \mid n$, that is, the number of chairs that Fermat can save is a divisor of $n$, smaller than $n$. Then it's not larger than $n/p$. But it is clear that he can save $n/p$: suppose Gauss starts on $1$ and Fermat tries to stop him from getting on multiples of $p$. The only case when Gauss can get on a multiple of $p$ by force is when he's on $i(2)(a_i+a_j)$ with $a_i, a_j$ the two possible multiples of $p$, but this number is a multiple of $p$ itself, so he can never start. Thus $f(n)=n-n/p$.

Now we have to mix in the $2$. Note that $f(2a) \geq 2f(a)$, since Gauss can look at the residues $0, 1, \ldots, 2a-1$ as the union of $0, 2, \ldots, 2a-2$ and $1, 3, \ldots, 2a-1$, and then play his strategy for $a$ in each subset (multiplying everything by $2$ and switching between subsets by calling out $1$). Then $f(2^cn) \geq 2^cf(n)$, where $n>1$ is odd. But, on the other hand, it's easy to see that Fermat can again save the multiples of $p$ for $p$ the smallest prime divisor of $n$, so $f(2^cn) \leq 2^cn-2^cn/p$.

Finally, $f(n)=n-n/p$ for all $n$, except for $n$ a power of $2$, when $f(2^c)=2^c$.

Can you solve my [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=54258]proposed problem[/url]?


The problem goes as follows.
Let $ a,b$ be some constant positive integers with $ gcd(a,b)\equal{}1$. For which values of $ N$ there is a pair $ (x,y)$ of non-negative integers such that we can write $ N\equal{}ax\plus{}by$?

I know that every number $ N>ab\minus{}a\minus{}b$ can be written in that form (this has been also posted before). But what can we say about the integers $ N<ab\minus{}a\minus{}b$? Is there a way to find which $ N<ab\minus{}a\minus{}b$ satisfy the conditions?

Thank you in advance for your help

[quote="t0rajir0u"]The only result I know in this direction is that exactly half of all such $ N$ can be written this way.  The values of $ N$ aren't easy to describe in general; what kind of a description are you looking for?[/quote]

First of all, thank you for your interest. The result to which you referred has to do with the fact that if $ N$ can be written in that form, then $ ab \minus{} a \minus{} b \minus{} N$ cannot be written like that (i do not know a proof about that, but I trust my friend who said that to me :P ).

concerning the description what I mean is whether there exists a formula which can give us the $ N$ which are less than $ ab \minus{} a \minus{} b$, which satisfy the condition (of course this formula will depend on $ a,b$). In addition if we have for example the number $ N \equal{} ab \minus{} 2a \minus{} 2b \minus{} 3$ how can we verify if it can be written as $ N \equal{} ax \plus{} by$ (probably here we just have to take some cases about $ a,b$) :?:

I hope I was more clear this time about my question :)


algebra
polynomial
Galois Theory
superior algebra
superior algebra unsolved

1. Find the splitting field L of x^8+8 over Q.
2. Find the Galois group of L/Q.

Thanks in advance. :)

1) Let $ a\equal{}e^{i\pi/4}\equal{}\dfrac{1\plus{}i}{\sqrt{2}}$. The roots of the polynomial are $ x_k\equal{}e^{i\pi/8}8^{1/8}a^k$. Let $ K\equal{}Q(a)$ ; $ a^2\equal{}i$ then $ i,\sqrt{2}\in{K}$ and
$ [K: Q]\equal{}4$. $ L\equal{}Q(u,a)\equal{}K(u)$ where $ u\equal{}e^{i\pi/8}8^{1/8}$ ; $ u^4\equal{}2i\sqrt{2}$ then $ [L: K]\equal{}4$ and $ [L,Q]\equal{}16$.
2) The roots $ (x_k)_{k\leq{8}}$ are the vertices of a convex octagone. If we want to prove that $ L/Q\equal{}D_8$ (that is true !) it remains to prove that the permutations $ (78)(16)(25)(34)$ and $ (12345678)$ are associated to  $ Q$-automorphisms of $ L$.

[quote="loup blanc"]1) Let $ a \equal{} e^{i\pi/4} \equal{} \dfrac{1 \plus{} i}{\sqrt {2}}$. The roots of the polynomial are $ x_k \equal{} e^{i\pi/8}8^{1/8}a^k$. Let $ K \equal{} Q(a)$ ; $ a^2 \equal{} i$ then $ i,\sqrt {2}\in{K}$ and
[/quote]
why is $ \sqrt {2} \in K$ ?
I think you should probably prove that $ \sqrt {2}$ can be expressed using $ a$ and the basic operations $ \plus{} , \minus{} ,*,/$ (becuase these are the operations of the field) no ?

[quote]
$ [K: Q] \equal{} 4$. $ L \equal{} Q(u,a) \equal{} K(u)$ where $ u \equal{} e^{i\pi/8}8^{1/8}$ ; $ u^4 \equal{} 2i\sqrt {2}$ then $ [L: K] \equal{} 4$ and $ [L,Q] \equal{} 16$.
2) The roots $ (x_k)_{k\leq{8}}$ are the vertices of a convex octagone. If we want to prove that $ L/Q \equal{} D_8$ (that is true !) it remains to prove that the permutations $ (78)(16)(25)(34)$ and $ (12345678)$ are associated to  $ Q$-automorphisms of $ L$.[/quote]

you mean $ \varepsilon_{u}$ and $ \varepsilon_{u^4}$ ?

in general, is it a good idea to guess what the Galois group would be by comparing the relations of the adjoined elements and their orders with those of some known groups ?
(if the degree of the polynomial is low I imagine there would not be so many groups to look at)

for instance in this case the diherdal group has the presentation:
$ \langle x, y \mid x^2 \equal{} y^2 \equal{} (xy)^n \equal{} 1 \rangle$

does this help in order for the computation of Galois groups to be more intuitive ?

Hi spx2.
1) $ \sqrt{2}\equal{}1/a\plus{}a$.
2) You are right about the presentation.
Let $ f,g$ be the $ Q$-automorphisms of $ L$ defined by:$ f: (a,u)\rightarrow (a^3,u)$ and $ g: (a,u)\rightarrow (\minus{}a,au)$.
$ f^2\equal{}g^8\equal{}(fg)^2\equal{}id$ and the Galois-group contains $ D_8$. Thus it is $ D_8$.


function
calculus
derivative
calculus computations

Let’s F be a function of two variables (x,y) which is continuous in a neighborhood of (x0,y0) while F(x0,y0) = 0 and suppose it would have continuous partial derivatives Fx ,Fy ,Fxx ,Fxy ,Fyy,Fyx in this neighborhood and Fy(x0,y0) # 0 .
Prove that F(x,y) has a unique answer as function f in a neighborhood of x0 such that F(x,f(x)) = 0 and f is continuously two times differentiable by the following formula :

f”(x) = - (Fxx(x,y)( Fy(x,y))^2 - 2 Fx(x,y) Fxy(x,y)+ Fyy(x,y)( Fx(x,y))^2 )/  (Fy(x,y))^3

implicit function theorem tells us that such an f exists

$ \frac{d}{dx}F(x, f(x)) \equal{} F_x (x,f(x)) \plus{} F_y (x,f(x)) f'(x)$

$ f'(x) \equal{} \minus{}\frac{F_x}{F_y}$

$ f''(x) \equal{} \minus{}\frac{F_y[F_{xx} \plus{} F_{xy} f'(x)] \minus{} F_x[ F_{yx} \plus{} F_{yy}f'(x)]}{F_y^2}$

$ \equal{} \minus{}\frac{F_y^2[F_{xx} \minus{} F_{xy} F_x] \minus{} F_xF_y[ F_{yx} \minus{} F_{yy}F_x]}{F_y^3}$

$ \equal{} \frac{F_y^2F_{xx} \minus{} F_{xy} F_x F_y \minus{} F_xF_y F_{yx} \plus{} F_{yy}F_x^2}{F_y^3}$

which is what you wanted


The math teacher I put in my Profile section in the USAMTS website said that he hadn't heard anything about the AIME yet - is that because people who qualify through USAMTS take the AIME II, or is this just a miscommunication? And if, for some reason, the fact that I qualified through USAMTS was misplaced or something, could I still take the AIME II once the problem was fixed?

I would say because you can take [i]both[/i]

You can take one AIME or the other, [i]but not[/i] both.

No, what I mean is this: I qualified to take the AIME through USAMTS. The teacher who I put down to send it to on the Profile page says he didn't get anything yet, or even hear from the AIME people at all - so I'm assuming there's been some sort of mix up. I was asking whether its because everyone who qualifies through the USAMTS gets the AIME II, not the AIME I.

Could I still use my qualification to take the AIME II if I miss the AIME I due to this potential mix up?

hope that clarifies my question

Yes, sorry, I meant either... :D
it was late...

You can take the AIME I

OR

the AIME II

but not both (i guess it would be xor?)

no matter how you qualify, you have two options: they have the two options to let people who can't take the AIME I take the AIME.

[quote="zatomics"]
Could I still use my qualification to take the AIME II if I miss the AIME I due to this potential mix up?

hope that clarifies my question[/quote]

Have your teacher contact the AMC right away.

I believe you will be able to take the AIME II in case of a mix-up like this, but the AMC sets this policy.

I went in to school today, and he said he had just received the test materials. Thanks for everyones replies, luckily it turned out okay!

The test was a lot of fun!!


its one that should be responsible for all activities in a forum

im used to it since i modded lots of forums before

as for why yufei is a mod, he is a good mathematician...great reason...and he started the petition for a new forum for internation contests!

great idea

[quote="bugzpodder"]why is billzhao the mod here?  how is the mod decided[/quote]

I can describe how I became the moderator of one forum here, and the co-moderator of another. In both cases, I started participating on the forum in question, and then after a while I sent Mr. Rusczyk (at last I am learning how to spell his name! I used to always cut and paste it) a private message asking him if he would permit me to be a moderator on each of those forums.

I imagine that billzhao became moderator because he was advocating a new forum so vigorously. I would like to see, in addition to information about the Canadian contests, information about contests in Britain, Australia, and even in the non-English-speaking world. I have read most posts to this forum (something I CAN'T say about Fun and Games) because it is fairly interesting.

It would be really nice for everyone to understand a bit more about every country's different math level and their contests that are given.

We could also see how other countries interact with math.

"as for why yufei is a mod, he is a good mathematician...great reason"
actually, this is a very stupid reason (if this is indeed why he was selected as a mod).
that makes no sense what so ever... there are a lot of good mathematicians here, he is not the only one...  besides, you dont need to be good at math to be mod...

look, he isnt even here to answer my questions...

[quote="Magnara"][quote="tokenadult"]Mr. Rusczyk (at last I am learning how to spell his name![/quote]

Aww, man, he is definately going to take away your modship now...:-)[/quote]

My son and I are both LOL at that reply.

[quote="probability1.01"]What's wrong with yufei as mod?  He's not killing your threads, is he? (or is he?  :twisted: )[/quote]

nothing much...  except he ignored my posts asking him and i am angry

-_-....you probably wanted to be mod urself...

...jealousy....

[quote="bugzpodder"]I wouldnt mind[/quote]

I think that bugzpodder is an eminent threat to AoPS.  We have intelligence that bugzpodder has intentions of obtaining nukular mod-destroyers.  In fact, we have found aluminum (soda?) cans that may well be used to enrich uranium.  Also, his toy trucks can be used as laboratories for producing anthrax.

(Obviously I'm just kidding)

Anyways, I think bugzpodder could be a decent mod, although I don't see why anyone would be so eager to do so.

I wonder why I never noticed this post ...

Mr. R asked me if I wanted to moderate and I replied yes. That simple.

Anyways, bugz, if you want to be a mod, just go ask, I wouldn't have a problem with it.

[quote="blahblahblah"][quote="bugzpodder"]I wouldnt mind[/quote]

What's the big deal?  This forum doesn't need multiple moderators, and he's clearly the most proficient of those that take part in the CMC contests, and plus, he advocated for it.[/quote]

there is no big deal.. whats your deal??  I said i wouldnt mind, but i never said i wanted to be a mod and kick yufei off.   besides, i was half joking when i said that.  i was just interested in why yufei was a mod and why he ignored me


x^4 +8*x*y+4* y^4 +4=0

Help plzzz if you know

(0,1) and (sqrt(2),0) are solutions. No other apparent ones seem to be there, but that is the solution they were probably looking for.

edit: (-sqrt(2),0) and (0,-1) as well

This should go under intermediate level math.

[quote="redsmartie"]
(x^{2} + 4{y^2})^2 + 4 = 0
(x^{2} + 4{y^2})^2 = 4

[/quote]

I believe that by moving the 4 to the right side, you should get -4, not 4.

Sorry to revive this but no-one has found an answer yet.
I can't seem to factor it.
This is the best I can do
$ x^4\plus{} 4(y^4\plus{}2xy\minus{}1)\equal{}0$

Can we solve this geometrically?

[quote="Fraenkel"]Sorry to revive this but no-one has found an answer yet.
I can't seem to factor it.
This is the best I can do
$ x^4 \plus{} 4(y^4 \plus{} 2xy \minus{} 1) \equal{} 0$

Can we solve this geometrically?[/quote]

Other than using methods that would be found in the conics section of volume 2, I don't know.

Also, I believe there are an infinite number of solutions.

[quote="redsmartie"]x^{4} + 8xy + 4y^{4} + 4 = 0
(x^{2} + 4{y^2})^2 + 4 = 0
(x^{2} + 4{y^2})^2 = 4
(x^{2} + 4{y^2}) = 2
x^2 + 4y^2 + (-2) = 0.

Someone else finish it off. I'm too lazy to.[/quote]

To redsmartie: [u]c3nar1us[/u] said it's $ \ x^{4} \plus{} 8xy \plus{} 4y^{4} \minus{} 4 \equal{} 0$ not $ x^{4} \plus{} 8xy \plus{} 4y^{4} \plus{} 4 \equal{} 0$

To [u]alanchou[/u]: I don't think there are going to be infinite numbers of solution..
What made you think that? Can you explain?

[quote="alanchou"]Well, the problem only has the equation, it doesn't even say to solve for x and y, and even so, the problem doesn't say integer solutions, or even real solutions for that matter.[/quote]
That is true... :lol:

are u sure that its x^4 + 8xy + y^4 - 4 = 0 ???
isnt it x^4 + 8 x^2 y^2 + y^4 - 4 = 0 ??
(i changed x and y beside 8 to x squared and y squared)


group theory
abstract algebra
function
superior algebra
superior algebra unsolved

If $ H$ is a subgroup of index $ 2$ in a finite group $ G$ and if $ x \in H$, prove that $ |x^H| \equal{} |x^G|$ or $ 2|x^H| \equal{} |x^G|$, where $ x^H$ is the conjugacy class of $ x$ in $ H$.

Do you  know that the number of elements in G has to be twice the number of elements in H and that H has to be normal in G?

Yes. But I dont know how to use this fact.

I just get a terrible proof,

first,define the function(just a function):
$ u: G \to G$,by $ g \to gxg^{\minus{}1}$
and
$ v: H \to H$,by $ h \to hxh^{\minus{}1}$
Define $ K_1\equal{}u^{\minus{}1}(x),K_2\equal{}v^{\minus{}1}(x)$,then $ K_1,K_2$ are two subgroups of $ G$
Then it is not hard to prove
$ |x^G|\equal{}\frac{|G|}{|K_1|},|x^H|\equal{}\frac{|H|}{|K_2|}$,
and:
$ K_2$ is a subgroup of $ K_1$

but,$ |x^G| \ge |x^H|$ means $ [K_2: K_1] \le 2$
Which proves what we need.


function
calculus
derivative
ratio
calculus computations

I'm on a 15 hour study binge at the min with little sleep...but ive been sitting here trying to figure something very elementary out and its just not sinking in :P

So Im trying to compute the MRTS (Marginal Rate of Technical Substitution)...dont need to know what that means, but you just need to know that for a function

 $ F(L,K) = [(aL^b + (1 - a)K^b)]^({}^1{}^/{}^b^)$

The marginal rate of technical substitution is given by MPL/MPk(the partial derivative of L)/ partial derivative of k)

if I do that I get

$ MPL = abL^(^b^ - ^1^)$
and
$ MPk = (1 - a)bK^({}^b{}^-{}^1^)$ if I divide them through, Im supposed to get $ (a/1 - a)*[(K/L)^1 ^ - ^ ^b^)$ i guess my question is, I understand if you flip the L/K to K/L you reduce the exponent of both by one, right? because x to the - 1 power is just 1/x. For some reason cannot get this. I can see that the b's cross out since they are a common factor, and I understand why A stays on the top and (1 - a) goes to the bottom (because only the variables are raised to the negative powers).. Its the part where I get $ [L^(^b^ -  ^1^]/[K^(^b^  -  ^1^)]$ somehow becoming $ [(K/L)^1^  -  ^B^)]$ that i get tripped up

Any help? Im exhausted and feeling really dumb :P

Please someone point out the obvious to me, im sure its something really really simply algebraically.

You mean why $ \frac{L^{b\minus{}1}}{K^{b\minus{}1}}\equal{}\left(\frac KL\right)^{1\minus{}b}$ ? :? Seems like you really need some rest :lol:. The key is that $ z^{\minus{}k}\equal{}(\tfrac {1}{z})^k$ and $ \frac{x^k}{y^k}\equal{}\left(\frac xy\right)^k$

Ok so i think i might have figured it out, The format L/K is not the "standard form" for this value because it is not very interesting. More interesting in economics is the K/L relationship (the capital to labor ratio), so in order to invert the fraction you just multiply both of their exponents by -1, right?

so the exponent b-1 becomes -1(B-1) = 1-B, right?

hah, thanks for the help to the poster above. I am having trouble concentrating right now, i've been up for almost 36 hours straight...and they say staying up for more than 24 hours has similar effects to being drunk over the legal limit...so maybe im just drunk on stupidity :P

Time for some rest, then, I guess..thanks


geometry
3D geometry
tetrahedron
sphere
integration
real analysis
real analysis unsolved

This is from a Brazilian Olympiad.
Find the maximum volume of a tetrahedron that can be subscribed in a ellipsoid of equation $ \dfrac{x^2}{9} \plus{} \dfrac{y^2}{16} \plus{} \dfrac{z^2}{25} \equal{} 1$
I have the solution here. But I can't understand. The man calculates the volume of the tetrahedron in a sphere of unitary radius. Then he finds the volume of this tethahedron subscribed in this sphere. Soon after, he does a linear transformation to transform the sphere into the ellipsoid. Like this:

T=  \[ \left(
\begin{array}{ccc}
3 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 5 \end{array}
\right)\]

Then he multiplies the volume of the tethahedron by the determinant of the transformation. Why?
I didn't understand the meaning of this transformation. Is that $ T: \Re^3\minus{}> \Re^3$ ?

If $ f: R^n\rightarrow R$ is continuous ad the multi-variables transformation $ X \equal{} T(U)$  has some regular properties(bijective,derivable,..)  you can write :
$ \displaystyle\int _{H} f(X).dX \equal{} \int _{T^{ \minus{} 1}(H)} |det(J_T(U)|f(T(U)).dU$ , where $ J_T$ is the Jacobian of $ T$ .
In your case the transformation $ (x,y,z) \equal{} T(u,v,w)$ is $ x \equal{} 3u,\ y \equal{} 4v ,\ z \equal{} 5w$ , so :
$ J \equal{} \left( \begin{array}{ccc} x'_u & x'_v & x'_w \\
y'_u & y'_v & y'_w \\
z'_u & z'_v & z'_w \end{array} \right) \equal{} \left( \begin{array}{ccc} 3 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 5 \end{array} \right)$

Because of $ T$ is  linear then a tethahedron is transformed in a tethraedron .. and so on ..
 :cool:


geometry
modular arithmetic
number theory
number theory proposed

[i]Author of the problem is Melkior Ornik (although, of course, it may be possible that there already exists a very similar or same problem author didn't know about). This was Problem 4 (Number Theory) for 9th grade on 1st Balkan Student Mathematical Competition 2008 - a competition 11 schools from Balkan area took part of during November 2008. All information about the competition can be found on http://www.mathlinks.ro/viewtopic.php?t=243614.[/i]

Find all natural numbers $ x$, $ y$ and $ z$ and a prime number $ p$, which is of the form $ 4k \plus{} 1$ ($ k\in\mathbb{N}$), for which $ 4^x \plus{} p^y \equal{} 15^z$.

Take a view modulo $ 4$:
if $ x\geq1$ we have $ 4^x \plus{} p^y\equiv 15^z\pmod{4} \Longrightarrow 1\equiv (\minus{}1)^z\pmod{4}$ (When $ x\equal{}0$, it is trivial seeing modulo $ 4$ that we'll have an absurd). That means we can write $ z\equal{}2c$.
Now we can see that $ p^y\equal{}(15^c\plus{}2^x)(15^c\minus{}2^x)$ give us $ 15^c\plus{}2^x\equal{}p^a$, and $ 15^c\minus{}2^x\equal{}p^b$, where $ a\plus{}b\equal{}y$.
Than we can see that $ p^a\minus{}p^b\equal{}2^{x\plus{}1}$ show us an absurd when $ b\geq1$ (because $ p\nmid2^{x\plus{}1}$), so $ 15^c\equal{}2^x\plus{}1$, and $ p^y\equal{}15^c\plus{}2^x\equal{}2^{x\plus{}1}\plus{}1$, which implies $ 15^c\equal{}2^x\plus{}1$. Of course $ x\geq4$, so seeing modulo $ 16$ we'll have $ c$ as an even number.
Now make $ c\equal{}2f$, and we see: $ (15^f\plus{}1)(15^f\minus{}1)\equal{}2^x$, and finally $ gcd(15^f\plus{}1,15^f\minus{}1)\equal{}2$ or $ 1$ show us that $ 15^f\minus{}1\equal{}1$ or $ 2$. The rest is easy.


geometry
probability
probability and stats

Hi, my problem has a computer programming application.
I'm writing a program involving generating millions of random numbers between a range and adding them together. (like rolling n dice and adding the values)
The program has 2 variables, a "mean" and a "variance".
So it generates numbers from mean-variance to mean+variance.
However, since it generates tons of these numbers, the program tends to be very slow. I was thinking there might be a formula to generate a sum if an area from 0 to 1 was generated and finding the x value where the area between mean-variance and x is the generated area (similar to finding a z-score of a normal curve given the probability of an event). The problem is, I'm not entirely sure where to start.
Does any one have a clue how to approach this?

I take it you are generating independent identically distributed random variables uniformly distributed on $ (\mu \minus{} a,\mu \plus{} a)$ for some mean $ \mu$ and $ a > 0$ (I'd rather not call it 'variance', because strictly speaking it isn't).  Let's call them $ X_k$, for $ 1\leq k\leq n$.  The formula you are looking for is given by the [url=http://en.wikipedia.org/wiki/Central_limit_theorem]Central Limit Theorem[/url], which in this case states that for large $ n$ we have approximately
\[ \sum_{k \equal{} 1}^nX_k\sim \mathcal{N}\left(n\mu , nVar(X)\right)\]
where $ \mathcal{N}\left(n\mu ,n Var(X)\right)$ denotes the normal distribution with mean $ n\mu$ and variance $ nVar(X)$, where $ Var(X)$ is the variance of each of your random variables.  The variance in this case is easy to compute, it's $ Var(X) \equal{} \frac {a^2}{3}$.  In summary, the sum of your random variables satisfies
\[ \boxed{\sum_{k \equal{} 1}^nX_k \sim \mathcal{N}\left(n\mu ,\frac {na^2}{3}\right)}\]
So all you need to do is generate just one normal random variable instead of summing all those millions of uniformly distributed ones, and it will give you the same result to a high degree of accuracy (the higher $ n$, the lower the relative error).


how many solutions are there for x in

a) $ x^{2}048-1=0$
b) $ x^{3}2-2^{3}2=0$

Are not there imaginary solutions using DeMoivre?

[quote="archimedes1"]I'm assuming you mean "how many real solutions"?
Also, remember to use brackets in LaTeX for exponents of more than one digit:
a)$ x^{2048}-1=0$

b)$ x^{32}-2^{32}=0$.

Anyways,
[hide]a) Since $ x^{2048}=1$, The only real solutions are 1 and -1.
For b), $ x^{32}=2^{32}$ so $ x$ can be 2 or -2.[/hide][/quote]

yes thats what i meant but can you factor for the second, or will the solutions stay the same

If the statements are equivalant, then yes they will have the solutions in each.

You can factor for the second and the first, but you'll end up with
a) $ (x-1)(x+1)($some unfactorable expression$ )$
b) $ (x+2)(x-2)($some unfactorable expression$ )$
The complex solutions are roots of the unfactorable expressions.

The complex solutions are simply found by using roots of unity.

[hide="a"]$ x^{2048}=1$, so thus $ x$ is a $ 2048^{th}$ root of unity.  The $ 2048^{th}$ roots of unity are of the form $ e^{\frac{\imath k\pi}{2048}}$, where $ k=0,2,4,6,...,4094$.  Part (b) can be solved similarly with $ 32^{nd}$ roots of unity.[/hide]


Does there exist a positive integer such that the product of its proper divisors ends with exactly 2001 zeroes?

Proper divisors of n are (at least I think they are) all divisors of n except 1 and n.

I proved this problem, but my solution i really long and ugly, and I'm sure there's shorter and nicer one..

A simple example is $2 \cdot 5^{2001}$.  We only need to show an example to prove the problem of existence.

It is easy (by considering the prime factorization of $n$) to construct all such numbers.  It is a very large set.

[quote="krustyteklown"]I'm not quite sure, but does $3^{2001} \cdot 10$ satisfy it?
Or maybe your proof was that there doesn't exist...

That has 2002*2 divisors. 2002 where it doesn't have 10 as a factor.

Out of the other 2002, the one that's the entire number itself doesn't count, but all other 2001 contribute a 0 to the product, so it ends in 2001 zeroes.[/quote]

It has 2002*2*2 divisors...

If $n$ has divisors $1=d_1<d_2<...<d_k=n$ then there is some conection between product of all of them and $n$ but i can't remember witch. I think is something like this \[d_1\cdot d_2\cdot \ldots \cdot d_k=n^{\frac{k}2}\] of course if $n$ isn't square so you can maybe get something from that, because i really don't understand what is this problem asking, when I read it I think that I know what should be done, but when I start reading other posts,  everything became unclear...

[quote="krustyteklown"]I'm not quite sure, but does $3^{2001} \cdot 10$ satisfy it?
Or maybe your proof was that there doesn't exist...

That has 2002*2 divisors. 2002 where it doesn't have 10 as a factor.

Out of the other 2002, the one that's the entire number itself doesn't count, but all other 2001 contribute a 0 to the product, so it ends in 2001 zeroes.[/quote]

Yeah, I didn't check all the cases, I thought that it was the same in others, so I didn't realise that there is such a number.

Thanks for helping.

M@re, yes, since $d_i \cdot d_{k-i+1}=n$, we can get that $d_1\cdot \ldots d_k=((d_1\cdot d_k) \cdot (d_2 \cdot d_{k-1}) \ldots (d_k \cdot d_1))^{\frac {1}{2}}=n^{\frac {k}{2}}$.


induction
combinatorics solved
combinatorics

In a  certain language there are only two letters, A and B, The words of this language  satisfy the following requirements:

(i) There are no words of length 1, and the only words of length 2 are AB and BB

(ii) A sequence of letters of length n > 2 is a word if and only if it can be created from some word of length less than n by the following construction: all letters A in the existing word are left unchanged, while each letter B is replaced by some word. (While performing this operation, the B's do not all have to be replaced by the same word.)

Determine for any n the number of words of length n.

Hint: Prove that a word can be obtained iff it starts with an even number of B-s (eventually 0) and it ends with B. The rest is easy.

Please post full solution Dzeta, because with such an hint it may be probable that no-one post again on the subject.

Pierre.

Sorry… I didn’t post the complete solution because it can be found in the 1999-2000 Olympiad Book and I don’t like to copy solutions.
We call string any finite sequence of a and b. As I said, I want to prove that a string is a word iff it ends with b and starts with an even (eventually 0, i.e. it stars with a) number of b-s. It’s clear that this is necessary: it is satisfied by ab and bb and whenever a transformation is applied to a string like this, the propriety remains true.
Conversely, use induction on n, the length of the string. For n=1 or 2 this is obvious. Now we prove it for n. First, note that any string with even length consisting only of b-s can be obtained (induction on the length). Take any word of length n which satisfies the 2 conditions. If it is aa…b, it comes from ab like a(a…b). If it is ab…b, there are two cases. If a is folowed by an even number of b-s, it arises as a(b…b) from ab. If a is folowed by an odd number of b-s, it arises as (ab)(…b) from bb. Finaly, if it is bb…ba…b (starts with 2k consecutive b-s), it comes from bb like (bb…b)(a…b).
The proof of the claim is complete. Let us show that the answer is pn=(2^n+2*(-1)^n)/3. For n=1 and n=2 this is true. We are done if we show that p(n+2)=pn+2^n. But this is clear because any word of length n+2 is either ax…xb, where x…x is any sequence of length n (2^n posibities) or bbx…x where x…x is any word of length n and we are done.

Don't be worry, because probably not all the participants can have this book, so it is a help for them....

Pierre.


logarithms
trigonometry
calculus
derivative
inequalities
inequalities proposed

Let $ ABC $ be an acute triangle . Prove that :
 $ tg\frac{A}{2} + tg\frac{B}{2} + tg\frac{C}{2} + tg\frac{A}{2}.tg\frac{B}{2}.tg \frac{C}{2} \geq \frac{10\sqrt3}{9} $

Not a pretty solution, but $\frac{d^2[\tan(x)]}{dx^2} = 2\sec^2(x)\tan(x) \ge 0 \forall x \in (0,\pi/2)$, and $\frac{d^2[\ln(\tan(x))]}{dx^2} = -\csc^2(x) + \sec^2(x) \ge 0 \iff x\in [\pi/4,\pi/2)$. Since $ABC$ is acute, at least two of its angles are at least $\pi/4$; hence we may equate them by Jensen. Hence, we need only consider $g(x) = 2\tan(x) - \tan(2x) - \tan^2(x)\tan(2x)$ for $\pi/4 \le x < \pi/2$. Since $g'(x) = -4(1+2\cos(2x))\sec^2(2x)\tan^2(x)$, it is evident that $g(x)$ decreases to $g(\pi/3) =  \frac{10\sqrt{3}}{9}$ then increases, as desired.

haha it is from a book written by professor Tran Phuong :P
i have the solution like mildoft using derivative
 :blush:

hei nttu why not post the imply ineq from that one :D

Mr Tran Phuong is not a proffesor and I also don't get this problem from his book  ;)
  My friend has a solution ; he changed the problem into :
 $ xy+yz+xz =1 $ , prove that : $ x+y+z +xyz \geq \frac{10\sqrt3}{9} $
   He changes $ z= \frac{1-xy}{x+y} $ and expands all .
  I'm trying to find out a nicer solution  ;)

[quote="Vasc"]A sharper inequality is the following:
If x,y,z are non-negative numbers such that $xy+yz+zx=1$, then

$x+y+z+3(2\sqrt{3}-3)xyz \ge 2$.[/quote]

    I think two problems can be solved by the same way .
  $ z= \frac{1-xy}{x+y} $
  The problem becomes : $ x^2[1-3(2\sqrt3 -3)y^2]+x[(6\sqrt3 - 8)y-2] +(y-1)^2 \geq 0 $
  $ \delta' =y[(6\sqrt3 -9)y^3- (12\sqrt3 -18)y^2 +(32-18\sqrt3)y -(6\sqrt3 -10)] <0 $  ;)

 Maybe you have other solutions , Vasc  :)

[quote="cezar lupu"]Vasc, I posted some time ago a inequality like this one. It can be done very easy by using Lagrange multipliers. ;)[/quote]

Don't use lagrange multipliers when it is unnecessary!  :mad:

my solution:

wlog let A=min{B,C,A}

so $tg(A/2)\leq\frac{1}{\sqrt{3}}$

we transform it into:

$2tgA/2+\frac{2cosA}{cosA/2(sinA/2+cos((B-C)/2)}\geq 2tgA/2+\frac{2cosA}{cosA/2(sinA/2+1)}$

let t=tgA/2 so $t\leq\frac{1}{\sqrt{3}}$

the ineq remains:

$\frac{1+t\sqrt{1+t^2}}{t+\sqrt{1+t^2}}\geq\frac{5\sqrt{3}}{9}$

f'(t)<0  t<1/sqrt3

$f'(x)=\frac{t(2t^2+t\sqrt{1+t^2}-2)}{(t+\sqrt{1+t^2})\sqrt{1+t^2}}$

$t\leq\frac{1}{\sqrt{3}}$

therefore f'(x)<0 then  f(t)>f(1/sqrt3) :D  :P  :ninja:

the folowing ineq is from VIET NAM UNIVERSITY ENTRANCE EXAM:

$tgA+tgB+tgC+tg\frac{A}{2}tg\frac{B}{2}tg\frac{C}{2}\geq \frac{28\sqrt{3}}{9}$

$2\sqrt{3}tg^2(\frac{A}{2})+4tg^3\frac{B}{2}+3\sqrt{3}tg^4\frac{C}{2}\geq\frac{13\sqrt{3}}{9}$

$\sum cos(\frac{A-B}{2})\geq cos\frac{1}{3}(A-PI/3)$ :P

good luck :rotfl:

[quote="nttu"]   The problem becomes : $ x^2[1-3(2\sqrt3 -3)y^2]+x[(6\sqrt3 - 8)y-2] +(y-1)^2 \geq 0 $
  $ \delta' =y[(6\sqrt3 -9)y^3- (12\sqrt3 -18)y^2 +(32-18\sqrt3)y -(6\sqrt3 -10)] <0 $  ;)

 Maybe you have other solutions , Vasc  :)[/quote]
I deduced this inequality
$\displaystyle x+y+z+3(2\sqrt{3}-3)xyz \ge 2$
by (n-1)Equal Variable Principle. But your method works if you assume that $y=min(x,y,z)$, $y\le 1/\sqrt3$, and write correct the
discriminant:
$ \delta' =y[(6\sqrt3 -9)y^3- (12\sqrt3 -18)y^2 +(33-18\sqrt3)y -(6\sqrt3 -10)]$ ,
$ \delta' =y(\sqrt3y-1)^2[(2\sqrt3-3)y-(6\sqrt3-10)] \le 0$.


function
vector
real analysis
real analysis unsolved

$ \begin{gathered}
  {\text{By considering the sequence of functions in }}C\left[ {0,1} \right]{\text{ defined by}} \\
   \\
  f_n \left( x \right) \equal{} nx \\
   \\
  {\text{for 0}} \leqslant x \leqslant 1/n,{\text{ }}f_n \left( x \right) \equal{} 1{\text{ for 1/}}n \leqslant x \leqslant 1,{\text{ show that }}C\left[ {0,1} \right]{\text{ is not locally compact}}{\text{.}} \\
\end{gathered}$

These functions are all in the closed unit ball of $ C[0,1].$ In fact, $ \|f_n\| \equal{} 1$ for each $ n.$ If the space were locally compact, it's closed unit ball would be compact; if its closed unit ball were compact, this sequence would have a convergent subsequence.

We can directly show that for $ m > n,\ \|f_m \minus{} f_n\| \equal{} 1 \minus{} \frac {n}{m}.$ If we'd like, we can write this as $ \|f_m \minus{} f_n\| \equal{} \frac {|m \minus{} n|}{\max(m,n)}.$

The problem with this is that for and $ n,$ if we have a subesequence $ f_{n_j},$ we can always find $ n_j$ large enough that $ 1 \minus{} \frac {n}{n_j} > \frac12.$ Therefore, no subsequence of this sequence can be a Cauchy sequence.

--

There's a theorem that says that a normed vector space is locally compact if and only if it is finite dimensional.


trigonometry
inequalities
rearrangement inequality
inequalities proposed

[quote="gemath"]Given triangle $ABC$ prove that
$\frac{\cos^{2}A}{\cos^{2}\frac{A}{2}}+\frac{\cos^{2}B}{\cos^{2}\frac{B}{2}}+\frac{\cos^{2}C}{\cos^{2}\frac{C}{2}}\ge 1$ :D[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=59649

I am sorry, gemath. Soarer's proof holds for acute triangle only.
I think, the zhaobin's method kills your first problem.
About zhaobin's method see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64933

[quote="gemath"]ok arqady there is the nice solution in your link, I still a similar inequality but I haven't solved it :
Given triangle $ABC$ prove that
$\frac{\cos^{2}A}{\cos^{2}\frac{B}{2}}+\frac{\cos^{2}B}{\cos^{2}\frac{C}{2}}+\frac{\cos^{2}C}{\cos^{2}\frac{A}{2}}\ge 1$  :)[/quote]
I think $\frac{\cos^{2}A}{\cos^{2}\frac{B}{2}}+\frac{\cos^{2}B}{\cos^{2}\frac{C}{2}}+\frac{\cos^{2}C}{\cos^{2}\frac{A}{2}}\ge \frac{\cos^{2}A}{\cos^{2}\frac{A}{2}}+\frac{\cos^{2}B}{\cos^{2}\frac{B}{2}}+\frac{\cos^{2}C}{\cos^{2}\frac{C}{2}}$by rearrangement inequality

[quote="gemath"]Really Hawk Tiger? you mean when $A\ge B\ge C$? Can you show clearly?[/quote]
whenever  $A\ge B\ge C$,$A\geq C\geq B$,it holds because of rearrangement inequlity.


calculus
integration
function
counting
distinguishability
real analysis
calculus computations

Let $ \dot{f}(x) \equal{} \begin{cases} 1 & \text{ if } x \ne 0 \\
35 & \text{ if } x \equal{} 0 \end{cases}$

and let $ \dot{g}(x) \equal{} 1$. Are these two function $ f(x)$ and $ g(x)$ indistinguishable, in the sense that if I integrate above two definition to get $ f(x)$ and $ g(x)$, are they going to be different? I also think there have same $ L_p$ norm.

Does integration make us lose local information because it is a summation?

Integration (Lebesgue) doesn't distinguish functions differing on a set of measure $ 0$. Riemann integration is a bit "pickier": it also doesn't care about sets of measure $ 0$ as long as both functions are Riemann integrable but it can sometimes distinguish the functions by refusing to integrate one of them. The reason is not as much in summation as in multiplication of values by the interval lengths (for Riemann), so if you change your function at just one point, there will be only one term in the Riemann sum affected by it and it tends to $ 0$ as the partitions become finer and finer.


trigonometry
function
real analysis
complex analysis
calculus
calculus computations

[quote="Nea"]Break apart by Maclaurin's formula:
$1)$ $f(x,y)=e^{x}siny$, $n=3$
$2)$ $f(x,y)=sinxsinhy$[/quote]

There are a few possible ways to answer this question, which is most useful depends on whether you are in a Calc I, Calc III, Real Analysis, or Complex Analysis class.

[b]i[/b].) If you are in a Calc I class:

$1)$ $f(x,y)=e^{x}\sin y = \left( \sum_{k=0}^{\infty}\frac{x^{k}}{k!}\right) \left( \sum_{j=0}^{\infty}\frac{(-1)^{j}y^{2j+1}}{(2j+1)!}\right)$
$=\left(1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots\right)\left(y-\frac{y^{3}}{3!}+\cdots\right)$
$=\left(1+x+\frac{1}{2}x^{2}+\frac{1}{6}x^{3}+\cdots\right)\left(y-\frac{1}{6}y^{3}+\cdots\right)$
$=y-\frac1{6}y^{3}+xy-\frac1{6}xy^{3}+\frac1{2}x^{2}y-\frac1{12}x^{2}y^{3}+\frac1{6}x^{3}y-\frac1{36}x^{3}y^{3}+\cdots$

but for $n=3$ we need only some of these terms, namely

$e^{x}\sin y= y-\frac1{6}y^{3}+xy+\frac1{2}x^{2}y+\cdots$

$2)$ $f(x,y)=\sin x\, \mbox{sinh}y= \left(\sum_{k=0}^{\infty}\frac{(-1)^{k}x^{2k+1}}{(2k+1)!}\right) \left(\sum_{j=0}^{\infty}\frac{y^{2j+1}}{(2j+1)!}\right)$
$= \left(x-\frac{x^{3}}{3!}+\cdots\right)\left(y+\frac{y^{3}}{3!}+\cdots\right)$
$=\left(x-\frac{1}{6}x^{3}+\cdots\right)\left(y+\frac{1}{6}y^{3}+\cdots\right)$
$=xy-\frac1{6}xy^{3}+\frac1{6}x^{3}y-\frac1{36}x^{3}y^{3}+\cdots$

but for $n=3$ we need only some of these terms, namely

$\sin x\, \mbox{sinh}y=xy+\cdots$

[b]ii[/b].) If you are in a Calc III or Real Analysis class: Recall that Taylor Theorem for bivariate functions is
\[f(x,y) = \sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{\partial^{n+k}f(x_{0},y_{0}) }{\partial^{n}x\partial^{k}y}\frac{(x-x_{0})^{n}}{n!}\frac{(y-y_{0})^{k}}{k!}\]
and since we are cited to use Maclaurin's formula, we will take $(x_{0},y_{0})=(0,0)$.

$1)$ For $f(x,y)=e^{x}\sin y$ we have that $\frac{\partial^{n+k}f(x,y) }{\partial^{n}x\partial^{k}y}=e^{x}\sin \left(y+k\frac{\pi}{2}\right)$
and hence that $\frac{\partial^{n+k}f(0,0) }{\partial^{n}x\partial^{k}y}=e^{0}\sin \left(0+k\frac{\pi}{2}\right) = \sin \left(k\frac{\pi}{2}\right)$

Note that $\sin \left(k\frac{\pi}{2}\right) = 0, 1, 0,-1, 0, 1,\ldots$ for $k=0,1,2,\ldots$ so we have

$f(x,y)=e^{x}\sin y = \sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{\partial^{n+k}f(0,0) }{\partial^{n}x\partial^{k}y}\frac{x^{n}}{n!}\frac{y^{k}}{k!}$
$= \sum_{n=0}^{\infty}\sum_{k=0}^{\infty}\frac{x^{n}}{n!}\frac{(-1)^{k}y^{2k+1}}{(2k+1)!}= \left( \sum_{n=0}^{\infty}\frac{x^{n}}{n!}\right) \left( \sum_{k=0}^{\infty}\frac{(-1)^{k}y^{2k+1}}{(2k+1)!}\right)$
$=\left(1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots\right)\left(y-\frac{y^{3}}{3!}+\cdots\right)$
$=\left(1+x+\frac{1}{2}x^{2}+\frac{1}{6}x^{3}+\cdots\right)\left(y-\frac{1}{6}y^{3}+\cdots\right)$
$=y-\frac1{6}y^{3}+xy-\frac1{6}xy^{3}+\frac1{2}x^{2}y-\frac1{12}x^{2}y^{3}+\frac1{6}x^{3}y-\frac1{36}x^{3}y^{3}+\cdots$

but for $n=3$ we need only some of these terms, namely

$e^{x}\sin y= y-\frac1{6}y^{3}+xy+\frac1{2}x^{2}y+\cdots$

[b]iii[/b] If you are in a Complex Analysis you might note that:

$1)$ $f(x,y)=e^{x}siny = \Re\left[e^{x+iy}\right] = \Re\left[\sum_{k=0}^{\infty}\frac{(x+iy)^{k}}{k!}\right] = \Re\left[ 1+(x+iy)+\frac{(x+iy)^{2}}{2}+\frac{(x+iy)^{3}}{3!}+\cdots\right]$

$2)$ $f(x,y)=\sin x\, \mbox{sinh}y= \Im\left[\cos (x-iy)\right] = \Im\left[\sum_{k=0}^{\infty}\frac{(x-iy)^{2k}}{(2k)!}\right]$

Thanx benorin. U're great in your answers and sorry for my incomplete question.


My students and I were working on this problem in class today:

If all the angles of a triangle were multiples of 12, how many different combination's of three angles exist?

Essentially, since most of my students aren't well versed in counting theory, we just did organized listing to solve this problem (19 is the answer).  However, one student asked if there was a formula to solve this.  Here is what I have come up with.

x + y + z = 180 then if you substitute a, b and c for x, y and z divided by 12 respectively, the problem becomes a + b + c = 15.  Even better, since a, b and c must be at least 1, the problem becomes a + b + c=12.  This problem is very easy if you are counting the number of ordered triples, but how do you look for all the possible a, b, c if order doesn't matter?

Any ideas?

So I finally came up with a viable solution:

There are 14C2 = 91 ordered natural numbered triples (x, y, x) that solve x + y + z =15.  One of these is (5, 5, 5) and only counted once.  Six more have doubles in them (1,1,13), (2, 2, 11)...(7, 7, 1) (got to be sure not to count 5,5,5) and these were counted 3 times.  Thus 91-1-18 = 72 the rest were distinct x, y, z and thus were all counted 6 times so there are 72/6 = 12 of those.  This shows there are 12+6+1 ways of doing this.

If you extend it to x + y + z = 30 (The problem I originally presented but multiples of 6).  It would be 29C2=406 total ordered pairs.  Minus 1 for 10, 10, 10, minus 13*3 for the doubles equals 366.  366/6 = 61.  So the answer would be 61+13+1=75.

Is there a better way to do this?

I know this is ancient history, but the question was never answered.

Anyway, this problem is the number of partitions of $n$ into exactly $k$ pieces.  Sometimes written as $p(n,k).$

This is often solved as a recursion, but I think it is quite tedius when $n$ is $15$ or $30$ as in your examples.  Your careful casework is a much better choice, in my opinion.  However, for the special case of $k=3,$ there's a simple algebraic formula.

The number of partitions of $n$ into exactly $3$ pieces is:
$p(n,3)=\left[\frac{n^2}{12}\right],$ where $[x]$ denotes the nearest integer to $x.$

$p(30,3)=\frac{900}{12}=\boxed{75}$

$p(15,3)=\left[\frac{225}{12}\right]=\left[18+\frac{3}{4}\right]=\boxed{19}$

Ref:  [url]http://www.research.att.com/~njas/sequences/A069905[/url]


What is it? What aids do we get to use? Is there a time limit?

In a master round, you are asked to prepare a presentation to a panel of judges on a specific topic (and you don't know what the topic is). You have a 30 minute limit to prepare a presentation and then 15 minutes to present, where 4 minutes will be dedicated to questons from the judges. After the presentation, the judges will score you on a scale of 1-25.
yes, you can use a calculator.

Its pretty hard to qualify for the USAMO as a Mathcounts student though.

Not really (by that I mean about 20 people can do it  :blush: ).  But this year it is easier since the AIMEs were so easy.  I liked last year better :(.  The proofs are about the level of, say, proving the law of sines.

So it's optional for the states? (My state didn't do it).

Maybe. (My state did it)

I know nats does it though.

Chapter and School is an impossibility

You guys probably have enough people interested in doing the masters round, while Utah. . . :blush:

I wouldn't want to do the masters round, I have stage fright...

Ugh, is [i]that[/i] what the masters' round is? I thought it was a tougher version of the countdown round or something... :what?:

That's what I was hoping, too, but unfortunately, no.

2001 State Masters Round right here, don't worry, this is from MathCounts [b][u]old[/u][/b] online site, so there's no copyright violation!

Edit: I can't open this myself...

Oh my. And this is a presentation? I hope if I get in at  nationals it's something I can [i]understand[/i] better than that so I don't get embarassed.

I tried a state level, took me about 15 minutes just to confuse my self.  I never got it.

Taking 15 minutes to confuse oneself...that's like listening to NPR on the way to school.  :P

[quote="probability1.01"]Generally something that you wouldn't have studied about.  It's basically testing your skills in communicating proofs, but the proofs themselves are not too hard.  If you can qualify or almost qualify for the USAMO you should be able to do the proofs.[/quote]

Extremely easy for you to say man.  ;)

What happens if you have no clue what to do??

The Masters Round is optional at states, as our state, Nebraska, didn't have one.

Another place you can get previous Masters Rounds are in the MATHCOUNTS books containing competitions from chapter, state, and nationals from previous years. I don't know where you can get them. I had a friend who somehow had the books.  :)

...although I don't know if seeing previous questions would do much good. Obviously, they wouldn't ask the same questions that they've asked in previous years, and looking at the old questions might just make you nervous if you don't know how to do them.  :)

Good luck at nationals to all of you peoples. I didn't make it. A pathetic 14th place. And in Nebraska too!  :(

Masters Rounds done at state are not made by MathCounts.  The people who run your State make them up.  This is because people got sick of how hard it was...I find them pretty easy to solve.  They are fun, but you have to be really creative.

The only masters round that exists now and is officially part of mathcounts is at Nationals.  It is A LOT harder at nationals then it is at state.

The best thing I could reccomend you do is practice some pretty hard proofs.  The AoPS book is a good way to start.  But don't look at the answer to the proof until you have at least spent 10 hours analyzing the problem.

Here is how I reccomend you going about a proof.

Read the question

Look to find absolutely everything that is given to you in the proof.  Sometimes it might not be so direct.

For example: This question from the COP Proof round caught RC-7th because he did't notice this one part:

[quote]Question 1:

$z_1,z_2,....z_5$ are non-negative real numbers that holds the following equation true:

[quote]Question 1:

$z_1,z_2,....z_5$ are non-negative real numbers that holds the following equation true:

$z_1+z_2+z_3+z_4+z_5 = 200$.

The maximum value, which is called $\mu$, is defined by $z_1+z_2$,$z_2+z_3$,$z_3+z_4$, or$z_4+z_5$. If $\mu$ lies between $\alpha$ and $\beta$ such that $\beta - \alpha = 2$, find, with fully written proof, the value of $x+y$ if $\mu - \beta + \alpha = \frac {x}{y}$. NOTE: $\alpha$ is the smaller number that is closest to the real value of $\mu$, meaning $\mu - \alpha$ is less than one.
[/quote].

The maximum value, which is called $\mu$, is defined by $z_1+z_2$,$z_2+z_3$,$z_3+z_4$, or$z_4+z_5$. If $\mu$ lies between $\alpha$ and $\beta$ such that $\beta - \alpha = 2$, find, with fully written proof, the value of $x+y$ if $\mu - \beta + \alpha = \frac {x}{y}$. NOTE: $\alpha$ is the smaller number that is closest to the real value of $\mu$, meaning $\mu - \alpha$ is less than one.
[/quote]

We are given that all $z_n$ are non-negative reals
It holds this true $z_1+z_2+z_3+z_4+z_5 = 200$
NOW THIS IS TRICKY

$\mu$, is defined by $z_1+z_2$,$z_2+z_3$,$z_3+z_4$, or$z_4+z_5$

If you look at it...this is telling you that:

$\mu=z_1+z_2=z_2+z_3=z_3=z_4=z_4+z_5$

And you could go on to solve the proof...
ANYWAY

Make sure you get all the given information from the proof!

Find out exactly what it is asking.

List out a set of possible methods, theorems that you could use to help you solve the proof

If you don't see an extremely direct method...Work cases.  It is not always the most efficient method of proving something, but it is correct.

Follow through with each idea looking for what might help you the most.

It may take a bit, but it eventually should all fall in place.

 :lol:

[quote="mysmartmouth"]If you have no clue, odds are no one else will either.  (That's assuming only the top ...?10?... make the Masters.[/quote]

The top 4 make it to the Master's Round.  Anyone can go and watch it.

[quote="Magixmaniak"]If you make it to the Masters Round at National Mathcounts but don't want to participate in it, aren't you allowed to give up your spot to the next highest-ranking person?[/quote]
Yes. (but who would want to? Its only 4 people competing for $8000?!?!)

$2000 actually (I think)

By the way, who won the 2003 Master's?

I dunno, but somebody from my school won it like 10 years ago...I plan to uphold this honor and reclaim the title for my school in Detroit!!! :rotfl:

Just kidding. I could never hope to do so.

[quote="Phelpedo"]2000 dollars actually (I think)

By the way, who won the 2003 Master's?[/quote]

I think it is 8000 dollars.

Btw people, do give up their spots in the Master's Round especially if they dont like speaking in public.

[color=darkblue][size=75][Moderator rcv's note:  Dollar signs removed, word dollar inserted, due to formatting problem.][/size][/color]

I think its $8000 (it was last year). Who is afraid of speaking in public?

Lots of people are afraid of speaking in public. I get a little skittish in front of a crowd like that. But that wouldn't stop my trying for $8000 if [i]I[/i] got in!

1. It is $2000 in fact (although this varies for year to year).
2. I'm horrible at and terrified of speaking in public.  But I did it anyway because I just figured, "Screw what all the other mathcountsers think of me."  I think the challenge mainly does not lie in the actual proof but rather how well you can explain it quickly.

I didn't know that you get money for winning the masters rounds. I think that a state master round from 2001 is on the site [url]http://yashar.biz/mathcounts[/url]. It's something on hexagons.


MATH

Maybe a relay or a guts or something. Some math.

I don't usually play computer/video games.

But I [i]always[/i] go onto AoPS. :)

GAHHHHHAHAHAHAHAHAHAAAAAAHHAHAHAHAH!
I was just about to post on the thread about best online game that if anyone says Runscape....

[quote="Nerd_of_the_Ages"]I was just about to post on the thread about best online game that if anyone says Runscape....[/quote]

Hehe RuneScape is one of the most popular games among 10-12 year olds. I used to play like 3 years ago but it got boring (and true online gaming is on the Xbox  :P ). I still log in on RS like once every other month to make sure my 5m networth account isn't deleted/hacked.

As for what I do on Saturday,

Tennis (depending on if parents say it's too cold; but remember, true tennis pros play year round)

Halo (when my parents are out)

Math (of course)

Some light reading.

Not sure if I'm misinterpreting that you're misinterpreting, but just clarifying, I meant that RC sucks.

Actually, RC in the CA bay area is way outdated. Most people play maple story. RC is like, so 8 years ago.

[quote="Nerd_of_the_Ages"]Actually, RC in the CA bay area is way outdated. Most people play maple story. RC is like, so 8 years ago.[/quote]

Wait and how old are these players? (Keep this on topic or else discuss this in the online games thread here).

Read, do Science Fair (just this week), AoPS, read, use the Internet, and mope around having nothing to do. Alas, I do play RuneScape  :oops: But not frequently.

[quote="D3m0n Shad0w"]But not frequently.[/quote]

Lies.

Heh, I used to play a lot of Runescape. Today, I basically spent the whole day doing math (a competition, and a bunch of old states).
Same for tomorrow.

I used to do a lot of goofing off.  Now I spend most of it doing homework *coughushistorycough*.  Once it's all over (that is, two months from now after APs), it's back to enjoying life for me.

[quote="D3m0n Shad0w"]At least I'm not hanging around the Internet playing chess, being outraged if I'm done a pawn.  :no:[/quote]

Lol, that's what my dad does. He's pretty good at chess.
7h3.D3m0n, I was referring to highschool people. But even now, very few middle school or elementary school people play it. This might be exclusive to the Fremont part of the bay area, though. I went to some other relatively nearby city once, and (disturbingly) all the computers were occupied - with people playing neopets and runescape.

Yeah, well now that I have my mom's computer back, I'm all starcraft on saturday (i think i said this before, but just trying to be on topic  :D )

I still remember when I used to play neopets on saturdays. Actually, neopets WAS a very good game, but it's gotten worse. Example? It used to say, "We will NEVER post advertisements on our website." Now just go to neopets.com and you'll see what it's like.  :wink:

But even then, I think I was considered a good player. Although being "pro" in neopets isn't much to brag about, you have to admit, some people suck crap at this game.

"SKILLZ! I've been playing for 5 months, and I have 2000 np! Omg, I pwn."
....

Actually, at the time I played the tried and true methods for accumulating np weren't there yet. If you had 1 mil, you were insanely rich. Now there are guides showing you how to get 1 mil in 1 month.

[quote="crazychinesemaniac"]Hey world! Just curious...So what you people enjoy doing on Saturday?????????

***I enjoy playing video games and if it's a nice day, like today, I like to go outside and play basketball or chill with friends***

Also does anyone here play Runescape????[/quote]

Like today, i usually go to chess tourneys and complete boring homework.

Haha here in Maryland mostly 12 year olds play RS, probably why JaGeX has grown partial to the exessive whining they do about their accounts.

Something to add to me Saturday list:

Usually there is a good movie on (yesterday it was Die Hard, the other day it was Black Hawk Down, or the Enemy of the State) so I watch that during the night.

I sometimes catch Stargate: SG-1 on TV too.

Some math(homework :P ), lots of reading, talking a LOT on the phone, a pinch of number theory(right now on palindromes.)

Oh, and a good movie(saw Pirates 3)  :))

I USED TO LOVE NEOPETS!!!!!

and yes, i sucked at it, too.. :)

and yeah.. neopets is dead.
for me, my obsessions were:
Pokemon
Neopets
Millsberry
BnB/Kartrider, (although I still want to play, but my parents banned it)
Miniclip/addictinggames
AoPS!!!

notice the maturity gap.. :D

on Saturdays I NORMALLY!!

wake up at 8:00 (later than usual)
watch TV until 9:30
eat breakfast
surf around AoPS/internet for an hour or so
go to parent's work and surf the internet there..
eat lunch at parents' work...
do some "educational" stuff.. (like my online class homework)
come home.

but NOOOOO today, I have a stupid concert that we have to go to a college about 2 hrs away from here.. D:
so i shall be playing violin for about 4 hours straight... -_-;;

on a good note, my sister's coming from college for Spring Break!! :D

Normally: Cramming Chinese homework for 15 minutes, then saying nahh...., and decide to go to Chinese school with half completed homework.

Ah, I love saturday....
wake up around 7:30...
go to karate
read random articles
listen to music on youtube
go to b&n (sometimes)
math
run
facebook
chat a bit on gmail
perhaps a game or 2 of chess (just started up again)
that's pretty much it.. unless there is some party which my parents make me go to

Math, tennis, homework, movies, AIM, programming, guitar, violin, and Xbox.


Let a1=1/2; a(n+1)=(2n-1/2n+3)an;
  Prove that a1+a2+...an<1
Can we replace 1 by a smaller number?

I think a1+a2+...an may not be < 1.
For your information, a1+a2+...a1000 = 1.24812780828757

No, $\displaystyle\sum_{n=1}^{1000} a_n=0.7496...$.

I believe that $1$ can be replaced by $\frac{3}{4}$.

Here is an explanation for $a_1+a_2+...a_n<1$.

The fraction $\frac{2n-1}{2n+3}$ is positive and less than $1$ for all $n\ge1$.  So, $a_n\le\left(\frac{1}{2}\right)^n$ for all $n\ge1$.

Since $\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^n=1$, the sum from $a_1$ to  $a_\infty$ (or any other series) is less than $1$.


A positive integer $n$ is given. In an association consisting of $n$ members work 6 commissions. Each commission contains at least $\frac{n}{4}$ persons. Prove that there exists two commissions containing at least $\frac{n}{30}$ persons in common.

Suppose the contrary. Suppose no two have at least n/30 in common. Then total members in commissions (inclusion-exclusion formula) > 6 * n/4 - 15 * n/30 = n. Contradiction.


The first number is 7, the second is 7+3, third is 7+3+4, the fourth is 7+3+4+4, the sixth is 7+3+4+4+3, 7+3+4+4+3+3, 7+3+4+4+3+3+4...

Sorry, I always have issues describing my ideas thrugh words.

M.Z.

[hide="Solution"]
The largest multiple of $7$ less than $2000$ is $1995$, by inspection.  Once the sequence reaches $1995$ it has $2 \times \frac{1995}{7} - 1 = 569$ terms, and then one more term is reached which adds either a $3$ or a $4$ (it does not matter which) for a total of $\boxed{570}$ terms.
[/hide]

[quote="boxedexe"]How many numbers less than or equal to 2000 that follow the sequence 7,10,14,18,21,24,28,...exist?[/quote]

Split into three sequences.

1. Numbers divisible by 7, ie. 7k for all positive integers k (7, 14, 21, ...)
2. Odd multiples of 7 + 3, ie. 7(2k-1) + 3 for all positive integers k (10, 24, 38, ...)
3. Even multiples of 7 + 4, ie. 7(2k) + 4 for all positive integers k (18, 32, 46, ...)

2000 = 7 * 285 + 5, so there are 285 terms in the first sequence
There are 143 odd multiples of 7 under 2000, and they all remain under 2000 when 3 is added
There are 142 even multiples of 7

Total number 570.


AMC
AIME
geometry
3D geometry
tetrahedron
rotation

"ABCD is a square side 6√2. EF is parallel to the square and has length 12√2. The faces BCF and ADE are equilateral. What is the volume of the solid ABCDEF?"

The solution online (kalva, of course) said to take EF central and parallel to ABCD and if you put two such solids together, you get a regular tetrahedron side length 12sqrt(2).

How do you put two of these solids together?

Okay, if this is the problem I think it is, you connect them on their square faces.  I wish I could show you a diagram, but I can't.  One of them is rotated with respect to the other.


The mean of ten numbers is 16.2. What is their sum?


function
abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Say we have this function that is an onto homomorphism (isomorphism right?). It goes from a ring R to a field F. We know that the kernel of f is an ideal by some theorem. But how could you go about proving it's a maximum ideal? I guess this is more of a question of how to go about proving an ideal is maximum.

Thanks.

suppose K is kernel of such a ring homomorphism which i'll denote by $\phi$ . then by the first  isomorphism theorem$R/ K$ is isomorphic to the image $\phi(R)=F$, since your homomorphism is surjective. therefore K is a maximal ideal

[quote="frogger34"]an onto homomorphism (isomorphism right?)[/quote]
No, an onto homomorphism is also called epimorphism (when the subject are algebraic structures). And isomorphism is an invertible homomorphism.