[hide]
It rearranges into
\[y^{4}-2y^{2}-4y+70=(x-12)^{2}\]
Now the left hand side is strictly greater than $(y^{2}-2)^{2}=y^{4}-4y^{2}+4$ since $(y-1)^{2}+32>0$. It's strictly less than $y^{4}$ when $y^{2}+2y-35=(y+7)(y-5)>0$, that is, when $y>5$. So if $y>5$, we must have
\[y^{4}-2y^{2}-4y+70=(y^{2}-1)^{2}\Rightarrow 4y=69\]
but this is not possible for integer $y$. Therefore $y=4$ or $y=5$. The first case clearly fails mod 4. The second case works, though (note the factorization in the $y^{4}$ bound) and we get $x-12=25$, or $x=37$. Since
\[37^{2}+5^{4}=1994\]
we're done.
[/hide]


Suppose Fred lives in a town where the bus comes exactly every 10 minutes. Assume he arrives at the bus stop at a random time.

(a) What is the distribution of how long Fred has to wait for the next bus? What is the average time that Fred has to wait?

(b) Given that the bus has not yet arrived after 6 min, what is the probability that Fred will have to wait at least 3 more min?

(c) Fred moves to another town, where this time the time until the next bus arrives is an exponential random variable with mean 10 min. Again, Fred arrives at a random time. What is the distribution of Fred's waiting time for the next bus? What is the average wait time?

(d) Explain why the answer to part c is not 5 min.

a) 0 to 10 minutes; average is 5 minutes
b)1/4

(c) His distribution is all positive reals.  Using the fact that the exponential distribution is memoryless I reduce the problem to finding the average time after one bus has arrived until the next comes, or 10 minutes.
(d) Because the exponential distribution is not the uniform distribution

I might have made a mistake in the reasoning for (c).

nope, you guys are correct  :lol:

care to elaborate on part c?


superior algebra
superior algebra solved

If $ A_1, A_2,...,A_n$ are nxn complex nilpotent matrices such that any two commute, than their product is the zero matrix.

It was posted a very long time ago [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=499&highlight=]here[/url].

Sorry, I did not know that. Anyway, I have a solution which does not use rank or endomorphisms, just a very nice identity. Sorry again for posting this.

harazi

Could you post this very nice identity ?

thx

The identity is:
  $ (x_1+x_2+...+x_n)^n-\sum {(x_2+...+x_n)^n}+\sum {(x_3+...+x_n)^n}-...=n! x_1\cdot x_2\cdot...\cdot x_n$.
   Here the sum $\sum {(x_2+...+x_n)^n}$ has n terms, the second one has $\frac{n(n-1)}{2}$ and so on.


[quote="i_like_pie"]The ratio of the lengths of the diagonals of a rhombus is $ 3: 4$. Find the ratio of the area of the rhombus to the area of the inscribed circle.[/quote]
[hide] Call the rhombus ABCD and the center of the circle O.  Notice that O is also the center of the rhombus, so both AC and BD go through O.  Now, call AC=6x and BD=8x.  So, we have that AO=3x and BO=4x.  So, we can say that AB=5x.  So, all of the sides of the rhombus are 5x.  Now, we have that the radius of the incircle of the rhombus the the length of the perpindicular from O to AB.  So, we drop this perpindicular OX.  Thus, we have that $ [ABC] \equal{} \frac {(AB)(BC)}{2} \equal{} \frac {(OX)(AC)}{2}$.  So, $ \frac {(AB)(BC)} \equal{} \frac {OX}{AC}$.  Thus, $ \frac {(3x)(4x)}{5x} \equal{} 2.4x$.  So, the radius of the circle is 2.4x and thus the area of the incircle is $ 5.76x^2 \pi$.  Now, the area of the rhombus is $ \frac {AC(BD)}{2} \equal{} 24x^2$.  So, this ratio is $ \frac {24x^2}{5.76x^2 \pi} \equal{} \boxed{\frac {25}{6 \pi}}$ [/hide]
EDIT: Oh, whoops!  Sorry, Farenhajt!

It would be more correct if some $ \pi$ appeared somewhere there... :)


Solving an geometry construction problem I get an quadratic equation $ ax^2\plus{}bx\plus{}c\equal{}0$, where $ a,b,c>0$ are known. Can I say that I can construct $ x$?

Yes.  There is a very nice explanation [url=http://www.cut-the-knot.org/Curriculum/Algebra/QuadraticRoots.shtml]here[/url].


trigonometry
real analysis
real analysis unsolved

[quote="Chen241290"][quote="BaBaK Ghalebi"]$f(x)=x^3+x+1$
$f^{-1}(x)=?$[/quote]
$\sqrt[3]{\frac{x-1}{2}+ \sqrt{\left(\frac{x-1}{2}\right)^2+\left(\frac{1}{3}\right)^3}}+\sqrt[3]{\frac{x-1}{2}- \sqrt{\left(\frac{x-1}{2}\right)^2+\left(\frac{1}{3}\right)^3}}$[/quote]
nice
but I`ve heard that this one has an aswer with $\sin$
now what do you think??


Please cast your poll.

Basic information:

There will be no difficulty difference between Test A and B. Only difference is their concepts and their date.

This probably will give more attention to test-takers.

Please choose your option carefully.

Thanks and post any of your comments.

If there are two tests, will the competition go any slower? What I mean is, will there be more delay between rounds, etc.

About a week or two.

Depends.

 :)

If you're bored, I can always make up shortlist for you to practice. :lol:

no. because for example, AMC is now a bit biased on index and all becuase of the second test... people can make up for what theyve made in the first test... i know COP does not involve around indices qualification, but it is not fair for someone who does well on the first test.

I passed the first test but would enjoy doing the second one as well.


I set up a class, and entered my three students who have set up accounts so far. For all three, I filled out boxes both for their real names, and for their user names. My class list shows two of them by their user name, and one by the real name and I can't switch anyone to make all the names consistant. I've tried dropping them and adding them back. I even tried setting up a new class. But I can't make the class diplay all the kids under either user name or real name. It doesn't matter with only 3 names. But I hope to get all of my students working on Alcumus in the fall. How can I get the display to show real names for all of them?

thanks!

Hello,

I'm not sure why, but the real names haven't been associated with the account. We'll look into this more on Monday. In the meantime, in your class list, simply click on the usernames and edit them. The real names should then be stored. Click the [ Help ] link listed in the class list header for more information.

That worked, thanks! Is there some way to change the order of the names to something other then ascending or descending "last on" dates?


geometry
ARML
calculus
integration
articles
algorithm
continued fraction

The smallest (in area) triangle with integer area and consecutive integer sides has sides 3,4,5. The next larger such triangle has sides 13,14,15. Compute the three sides of the next larger such triangle, given that the "middle" side is a multiple of 4.

What's an ARML #18?

[hide="Answer"]I got $ (51,52,53)$ by doodling with Pythagorean Triples for several minutes  :maybe: [/hide]

[b]Edit:[/b] [i]Here's a more legitimate approach[/i]
[hide="Solution"]
Using $ (4n \minus{} 1,4n,4n \plus{} 1)$, we get that the area is $ 2n \sqrt {12 n^2 \minus{} 3}$ by Heron's. We want to make the inside an integer. Thus, $ 12 n^2 \minus{} 3 \equal{} h^2$ for some $ h$. Clearly, $ 3|h$, so let $ k \equal{} h/3$, and we have $ 4 n^2 \minus{} 1 \equal{} 3 k^2$.

Now, we go hunting in the continued fraction representation of $ \sqrt {\frac {3}{2}}$, and after a couple iterations, find the convenient fraction $ \frac {13}{15}$. This suggests that $ (n,k) \equal{} (13,15)$ will work, and indeed, that is the case.

$ n \equal{} 13 \implies (4n \minus{} 1, 4n,4n \plus{} 1) \equal{} \boxed{(51,52,53)}$, which is believably low. If you have extra time, you can even check the other 9 values of the middle term.

If, for fun, we take the next term in the CF series, it doesn't work, but the one after that produces $ n \equal{} 181 \implies (723,724,725)$[/hide]

I meant #8...
That's the right answer.
I just realized that the solution to this problem is in the ARML marathon thread (problem 11), so sorry for posting this twice- I didn't find it when I first searched for it. :(

Haha, I don't know why that didn't occur to me, my bad

[b]Challenge:[/b] Prove that the numerators of alternating terms in the continued fraction series for $ \sqrt {3}/2$ provide the series of valid $ n$ values (ie, 1/4 the middle term) for the problem (ie. area is integral and 4 divides the middle term)

[quote="The Zuton Force"]

Now, we go hunting in the continued fraction representation of $ \sqrt {\frac {3}{2}}$, and after a couple iterations, find the convenient fraction $ \frac {13}{15}$. This suggests that $ (n,k) \equal{} (13,15)$ will work, and indeed, that is the case.
[/quote]

What in the world?  Where did you get $ \sqrt {\frac {3}{2}}$ from?  And what is this continued fraction thing?

Irrational numbers can be expressed as continued fractions.  For example:

$ \phi \equal{} (1 \plus{} \sqrt {5})/2 \equal{} \cfrac {1} {1 \plus{} \cfrac {1} {1 \plus{} \cfrac{1} {1 \plus{} ...}}}$

But I'm not sure where he got $ \sqrt {3}/2$.

$ \frac{13}{15}$ is not a convergent of $ \sqrt{ \frac{3}{2} }$.  You meant $ \frac{ \sqrt{3} }{2}$.  In any case, what we're really looking at is solutions to $ x^2 \minus{} 3k^2 \equal{} 1$ where $ x \equal{} 2n$ (i.e. $ x$ is even).  The Zuton Force is alluding to a rather beautiful technique used to solve problems of this type, which are called [url=http://en.wikipedia.org/wiki/Pell%27s_equation]Pell's equations[/url].  It turns out, rather magically, that the solutions to $ x^2 \minus{} dy^2 \equal{} 1$ are deeply intertwined with the continued fraction representation of $ \sqrt{d}$.

So after reading through the AoPS discussion and Wikipedia article, that makes sense.  But how are you supposed to find the continued fraction without a calculator on ARML?  Is there a better way to do this besides just plugging in values?

[url=http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Continued_fraction_expansion]The following algorithm[/url] can be efficiently done in tabular format by hand.

I think pi^2/6 meant problem #8 on the ARML individual in 1990. ARML gave another solution that does not involve Pell's equation or continued fractions (I doubt they'd expect most high schoolers to find such a solution in under 10 minutes).

Let the sides of the triangle be $ 4n \minus{} 1$, $ 4n$, $ 4n \plus{} 1$. Then the semiperimeter is $ 6n$ and the area, given by Hero's formula, is $ A \equal{} \sqrt {6n(2n \plus{} 1)(2n)(2n \minus{} 1)}\ \Rightarrow \ 2n\sqrt {3(2n \plus{} 1)(2n \minus{} 1)}$

Since the area must be integral, $ (2n \plus{} 1)(2n \minus{} 1) \equal{} 3k^{2}$, but $ 2n \plus{} 1$ and $ 2n \minus{} 1$ are relatively prime (they are both odd and cannot share any factor greater than 2). This implies that one number must be an odd square that is not a multiple of 3, and the other must be 3 times and odd square.
Plugging in $ 1 \equal{} 2n \minus{} 1\ \Rightarrow \ n \equal{} 1$ yields $ (3,4,5)$, and plugging in $ 25 \equal{} 2n \minus{} 1 \Rightarrow n \equal{} 13$ yields $ \boxed{(51,52,53)}$

Not bad for a #8 on arml  :lol:

I forgot Pell's so here's what I did:
As before, using Heron's, we have 4a^2=3p^2+1. P must be odd, so we sub in 2n+1 for p and get 3n^2+3n+1=a^2. Now, 3n(n+1) is always even, hence, a must be odd. So we sub in a=2m+1. Simplifying gives 3n(n+1)=4m(m+1). Since  m(m+1) is even, n(n+1) must be divisible by 8, so we find the smallest is m=6, n=7. Subbing back in, the sides are 51,52,53.


geometry
geometric transformation
rotation

Okay as requested by math92- Stupid Game 2

                      Rules:
1.Sign up. plokoon 51, ToBe, Forsaken, and gr8sk8r are in this round already unless they don't want to be.
2.I'll say a topic- say wookies, you name one each turn.
3. As far as I can understand the person who repeats a topic name or can't think of one will be out  (of the round)    :)
4. WARNING: this game is stupid.  :fool:

Oh yes. Instead of writing I cant think of any more, just put this emiticon: :surrender:

Yup Yup.. lets start? whats the first topic? wookies?

I don't think that there are that many names of Wookies that we know.  I think a better topic would be planets or maybe jedi again.

lets just make up some Wookies names, there is an entire population of Wookies that not even George Lucas know all of them :D

[quote="shobber"]can you give me a more detailed explanation of this game? it seems to be interesting.[/quote]

Here is a more detailed explination.
I'll name a topic for example wookies.
The order that you list the topic(or in this case wookies) is the order that you signed up in
The first person would go and say Chewbacca or another wookie
We keep going and if you rename Chewbacca or can't think of another wookie you are out.
When you have two people left they just battle it out.
So would you like to play  :?: :)

[quote="shobber"]can you give me a more detailed explanation of this game? it seems to be interesting.[/quote]the thing you have to do is name names on the topic or see Stupid Game detail

oops i didn't see that you already answer the question :blush:

Let's start! The first topic will be[hide]planets[/hide]

earth (i just like starting with.. yeah ill just not explain that)

Jupiter

What happens if someone says something that's not what the topic is?

I guess they are out?

And, should i post the next planet? or do we let the others post on until we go through a rotation?

oh oops... sorry...  ok!  Round 5!

planets named: Naboo, Tatooine, Alderaan, Hoth, Mustafar, Bespin, Dagobah, Coruscant, Yavin, Geonosis, Ando, Endor, Ilum, Utapau

hmm... I'll say Kessel!  (I'm trying to hit all the planets from the original trilogy first ;) - Han did mention Kessel)

[quote="CLee"]  When Luke made HIS lightsaber, he didn't go to Ilum [/quote]

I thought in the movie Obi-Wan gave him Anakin's old one. By the way, how do you switch lightsaber colors? In one movie Anakin had a blue one. In the next movie he has a red one. And how did Obi-Wan get ahold of Anakin's lightsaber? :huh: just wondering. :lol:

Luke lost the lightsaber Obi-Wan gave him on Bespin when his father cut off his right hand.  As for when Obi-Wan gets Anakin's lightsaber, well:
[hide]
He gets it after defeating Anakin on Mustafar in Episode III.[/hide]

Luke made a new one on Tatooine.  This one was green and appeared in RotJ.  You can't just switch colors.  The color is determined by the focusing crystal used for the lightsaber.  Obviously Anakin's lightsaber was different (if you're talking about the difference between Episodes III and IV).[/hide]

[quote="CLee"] plokoon51 is another SW book freak like me (well, comic books count, too)... um, right?[/quote]

correct. i'm more of a fan of the jedi apprentice and jedi quest series (Old Republic...good times :D )

i'll go with a planet that is only mentioned in books ( i doubt its been mentioned in movies). Iridonia: anyone knows which famous movie character was from there?

btw, i'm going to be away from the internet for 3 weeks so i won't be able to play any more... :(

slick.  the website barely covered that - yes, I'm so ashamed of myself, I actually HAD to look that up.  It's Darth Maul, but I'm not very good with pre-TPM.  I don't know why.  It just doesn't attract me much.  My avatar is post-RotJ...  New Republic... good times... :D  :D  :D haha.  You guys know what my lazy abbreviations are, right?  Sheesh, typing RotS just doesn't seem right...  Can ANYONE guess my avatar???  Oh and plokoon51, you're welcome to participate in Stupid Game, uh (1), I guess.  C'mon, my avatar's husband appears in the movies...  And I heard she gets a really brief seen in RotJ special edition DVD or something...  And she is the best SW character EVER...  Sorry.  I'll stop that now.

EDIT: oh, sorry, plokoon51, that really sux.  Anyways, math92 and I will keep posting (right, math92??? - don't forget about Stupid Game (1), either... )

i'll say Galantos.  it's were some wookies live.

[quote="mathclass"][quote="plokoon51"][quote="CLee"]Can ANYONE guess my avatar??? [/quote]

isn't that luke's wife or someone? (mara jade) i thought she was at first trying to kill luke or something...and aren't jedi not supposed to marry...(probably some old republic rule)[/quote]

yeah, but I don't think she looks as pretty in that avatar. she looks too skinny... she's supposed to be a really beautiful red-head.[/quote]

Yes, it's Mara Jade Skywalker.  You probably know the motivation behind her wanting to kill Luke about five years after RotJ.  They married ten years later.   :P Jedi are forbidden attachment.  Fine line there, since you'd have to wonder about Luke and Mara...  Luke *almost* crosses the line in [u]Vision of the Future[/u] ;) - if you've read it, it's the part when he loses mental contact with Mara and thinks she's dead...  Mara thinks the same thing, and then (haha) they realize they're being kinda stupid...   :D  :D  :D I still laugh when I think about that.

mathclass, yeah, I don't think they should have chosen Shannon to model for Mara on the trading cards.  Totally not fiery enough, and she doesn't represent Mara's character.  The only thing the share is gender and hair color.  lol.  I'm still trying to see if her eyes are even green...

math92, let me guess: NJO series, [u]Force Heretic I: Remnant[/u] by Shane Dix.  Yeah?  Either that or it's from one of the Star Wars Roleplaying Adventure thingys.

Round 6: I'll say... Kamino

So far, the planets named are: Naboo, Tatooine, Alderaan, Hoth, Mustafar, Bespin, Dagobah, Coruscant, Yavin, Geonosis, Ando, Endor, Ilum, Utapau, Kessel, Iridonia, Galantos, Kamino

gee, math92, speaking of Wookiees, there's one planet a zillion times as obvious as Galantos... ;)

yup.  I'll say Kashyyyk.  Someone said the last y is not pronounced.  I disagree.  Isn't it pronounced like Ka-shy-yik?

I pronounce it Ka-shay-ik
I can't think of anymore names so i guess you win CLee :surrender:
 :clap2:
i get  :winner_second:

o well.  stupid game 2 over...  :winner_first: with reading, I can come up with all sorts of names... and then struggle to remember where they came from.


function
limit
geometric series
algebra unsolved
algebra

Let $ a \in \mathbb{R}, 0 < a < 1,$ and $ f$ a continuous function on $ [0, 1]$ satisfying $ f(0) \equal{} 0, f(1) \equal{} 1,$ and

\[ f \left( \frac{x\plus{}y}{2} \right) \equal{} (1\minus{}a) f(x) \plus{} a f(y) \quad \forall x,y \in [0,1] \text{ with } x \leq y.\]

Determine $ f \left( \frac{1}{7} \right).$

If $ a\not \equal{}\frac 12$ function is not continiosly. If $ a\equal{}\frac 12$, then for continiosly function $ f(x)\equal{}x$.

since the left part of the equation is symmetric , so the right part must be symmetric too :

then :

$ (1\minus{}a)f(x)\plus{}af(y)\equal{}af(x)\plus{}(1\minus{}a)f(y)$

$ \longrightarrow$   $ a\equal{}1/2$  or $ f(x)\equal{}f(y)$

then $ f(x)\equal{}x$

:)

You dropped the assumption $ x \leq y$ by switching the variables.

My solution and answer are different than those previously mentioned, so hopefully if I have an inaccuracy someone can tell me what I did wrong.
[hide]
We first write the given equation as:
\[ f(x \plus{} y) \equal{} (1 \minus{} a) f(2x) \plus{} a f(2y)\]
with $ x,y\in [0,\frac {1}{2}]$. With $ x \equal{} 0$, we have $ f(x) \equal{} af(2x)$. From here, we can easily show inductively that $ f(\frac {1}{2^k}) \equal{} a^k$. We can again rewrite the given equation again as:
\[ f(x \plus{} y) \equal{} \frac {(1 \minus{} a)}{a} f(x) \plus{} f(y)\]
with the same previous constraints. Now, consider the sequence $ s_1 \equal{} 1/8$, $ s_{k \plus{} 1} \equal{} s_{k} \plus{} \frac {1}{8^{k \plus{} 1}}$. We make two observations: Firstly, since $ {s_n}$ is a geometric series, that $ \lim_{k\rightarrow \infty}s_k \equal{} \frac {1}{7}$. Also, since $ s_i > 0$ for all $ i > 0$, we have:
\[ s_k \equal{} s_{k \minus{} 1} \plus{} \frac {1}{8^k} > \frac {1}{8^k} > \frac {1}{8^{k \plus{} 1}}\]
We now have:
\[ f(s_{k \plus{} 1}) \equal{} f(s_k \plus{} \frac {1}{8^{k \plus{} 1}}) \equal{} (\frac {1 \minus{} a}{a})f(\frac {1}{8^{k \plus{} 1}}) \plus{} f(s_k)\]
Since $ f(\frac {1}{2^k}) \equal{} a^k,$ we have $ f(\frac {1}{8^{k \plus{} 1}}) \equal{} a^{3k \plus{} 3}$, and consequentially:
\[ f(s_{k \plus{} 1}) \equal{} a^{3k \plus{} 2}\cdot (1 \minus{} a) \plus{} f(s_k)\]
Now, we have:
\[ f(s_{k \plus{} 1}) \minus{} f(s_1) \equal{} \sum_{i \equal{} 1}^k f(s_{i \plus{} 1}) \minus{} f(s_i) \equal{} \sum_{i \equal{} 1}^k a^{3i \plus{} 2}\cdot (1 \minus{} a)\]

\[ \equal{} (1 \minus{} a)a^5 \cdot (\frac {1 \minus{} a^{3k}}{1 \minus{} a^3}) \equal{} (1 \minus{} a^{3k})\cdot(\frac {a^5}{a^2 \plus{} a \plus{} 1})\]
Hence,
\[ f(s_{k \plus{} 1}) \equal{} (1 \minus{} a^{3k})\cdot(\frac {a^5}{a^2 \plus{} a \plus{} 1}) \plus{} f(1/8) \equal{} (1 \minus{} a^{3k})\cdot(\frac {a^5}{a^2 \plus{} a \plus{} 1}) \plus{} a^3\]
Finally, we have:
\[ f(\frac {1}{7}) \equal{} f(\lim_{k\rightarrow \infty} s_k) \equal{} \lim_{k\rightarrow \infty} f(s_k) \equal{} \lim_{k\rightarrow \infty} (1 \minus{} a^{3k \minus{} 3})\cdot(\frac {a^5}{a^2 \plus{} a \plus{} 1}) \plus{} a^3\]

\[ \equal{} \frac {2a^5 \plus{} a^4 \plus{} a^3}{a^2 \plus{} a \plus{} 1}\]
since $ f$ is continuous.

[/hide]

[quote="Rust"]If $ a\not \equal{} \frac 12$ function is not continiosly. If $ a \equal{} \frac 12$, then for continiosly function $ f(x) \equal{} x$.[/quote]
I repeat my answer.

Why is $ f$ not continuous if $ a\neq\frac{1}{2}$?

I give prove in this forum for function
$ f(bx\plus{}(1\minus{}b)y)\equal{}af(x)\plus{}(1\minus{}a)f(y)$ if f is not const and $ a\not \equal{}b$, then f can not be continiosly.

[quote="orl"]Let $ a \in \mathbb{R}, 0 < a < 1,$ and $ f$ a continuous function on $ [0, 1]$ satisfying $ f(0) \equal{} 0, f(1) \equal{} 1,$ and
\[ f \left( \frac {x \plus{} y}{2} \right) \equal{} (1 \minus{} a) f(x) \plus{} a f(y) \quad \forall x,y \in [0,1] \text{ with } x \leq y.\]
Determine $ f \left( \frac {1}{7} \right).$[/quote]
$ P(0,x)\Rightarrow  f( \frac{x}{2} )\equal{}af(x)$
$ P(2x,4x) \Rightarrow  af(3x)\equal{}(1\minus{}a)af(2x)\plus{}a^2 f(4x)\equal{}(2\minus{}a) f(x)$
$ P(x,3x) \Rightarrow  f(x)\equal{}af(2x)\equal{}(1\minus{}a)af(x)\plus{} a^2 f(3x)$
$ \equal{}(1\minus{}a)af(x)\plus{}a(2\minus{}a) f(x)\equal{}a(3\minus{}2a)f(x)$
$ \Leftrightarrow f(x)(2a\minus{}1)(a\minus{}1)\equal{}0$
It gives that if $ a \ne \frac{1}{2}$ then  $ f(x)\equal{}0 \forall  0 \le x \le \frac{1}{4}$ which is wrong because :$ f(\frac{1}{4})\equal{}a^2 f(1)\equal{}a^2 \ne 0$
$ \Rightarrow a\equal{}\frac{1}{2}$
$ \Rightarrow f(\frac{1}{7})\equal{}\frac{1}{7}$
Have I gotten mistake somewhere ?

To rust: Does your problem have some condition as orl's ?

[quote="Vincent Gilbert"][quote="orl"]
Have I gotten mistake somewhere ?

To rust: Does your problem have some condition as orl's ?[/quote][/quote]
No, have not mistake. $ a\equal{}1\to f\equal{}const$.
Posted before similar problem and I prove if $ f(bx\plus{}(1\minus{}b)y)\equal{}af(x)\plus{}(1\minus{}a)f(y), 0<a,b<1$ and $ f\not \equal{}const$, then $ f$ can not be continiosly, when $ a\not \equal{}b$.

[quote="Rust"]I give prove in this forum for function
$ f(bx \plus{} (1 \minus{} b)y) \equal{} af(x) \plus{} (1 \minus{} a)f(y)$ if f is not const and $ a\not \equal{} b$, then f can not be continiosly.[/quote]

I don't get this. Can you elaborate a little more?


calculus
integration
trigonometry
function
calculus computations

Could you help me with

integral {Sqrt.(1+sin(alpha))}.d(alpha)

[hide="Hint"][quote="Jean-Louis"]Could you help me with
$ \int{\sqrt{1\plus{}\sin\alpha}}\ d\alpha$[/quote]
$ \sqrt{1\plus{}\sin\alpha}\equal{}|\sin\frac{\alpha}{2}\plus{}\cos\frac{\alpha}{2}|$[/hide]

[hide="Answer"]$ \int\sqrt{1 \plus{} \sin x}\,dx \equal{} \int \sqrt{\sin^2\frac{x}{2} \plus{} \cos^2\frac{x}{2} \plus{} 2 \sin \frac{x}{2} \cos \frac{x}{2}}\,dx$

$ \equal{} \int \sqrt{(\sin \frac{x}{2} \plus{} \cos \frac{x}{2})^2}\,dx \equal{} \sqrt{2}\int\sin\left(\frac{x}{2} \plus{} \frac{\pi}{4}\right)\,dx \equal{} \minus{}2 \sqrt{2} \cos\left(\frac{x}{2} \plus{} \frac{\pi}{4}\right) \plus{} C$[/hide]

To [b]carcul[/b]: You must to consider $ 2$ cases: $ \sin\frac {x}{2} \plus{} \cos\frac {x}{2}\geq0, < 0$


prove that if0 <x< (pi\2)
then
$\sec^{6}x+\csc^{6}x+\sec^{6}x.\csc^{6}x\geq 80$

[quote="ashrafmod"]prove that if0 <x< (pi\2)
then
$\sec^{6}x+\csc^{6}x+\sec^{6}x+\csc^{6}x\geq 80$[/quote]
It's not true, for x=pi/4 the result is 32.


Let $ n$ be a positive integer. Does there always exist a positve integer $ a$ ( $ a \in \{p,p^k,2p^k\}$) such that $ n$ is a primitve root in $ Z_a$?Here, $ p$ is prime.

Any idea or any knowledge on this are more then welcomed. :lol:

Artin's conjecture addresses the case of $ a\equal{}p$.
See the following review for further details:
M. Ram Murty. [url=http://www.mast.queensu.ca/~murty/mi.dvi]Artin's conjecture for primitive roots[/url], Math. Intelligencer, 10 (4) (1988) 59-67.


factorial
inequalities proposed
inequalities

Hi,

Show that for every natural $ n > 0$, we have :

[img]http://www.texify.com/img/%5CLARGE%5C%21%5Cfrac%7B1%7D%7B%5Csqrt%5Bn%21%5D%7Bn%2B1%7D%7D%2B%5Cfrac%7B1%7D%7B%5Csqrt%5Bn%5D%7Bn%21%2B1%7D%7D%5Cgeq%5Cfrac%7Bn%21%7D%7Bn%5E%7Bn-1%7D%7D.gif[/img]

[hide]first term should be very very close to 1, right-hand side should be much smaller than 1[/hide]

divide both sides by $ n$.
check the trend of LHS and RHS.and so by going on to $ n\to infinity$.
compare it to $ n\equal{}1$.
i think this would work.

$ n!\sqrt[n!]{(n\plus{}1).1...1}\leq n\plus{}1\plus{}(n!\minus{}1).1\equal{}n!\plus{}n$ (AM-GM). On the other hand $ n\sqrt[n]{(n!\plus{}1).1...1}\leq n!\plus{}n$ (AM-GM). Hence $ \frac{1}{\sqrt[n!]{n\plus{}1}}\plus{}\frac{1}{\sqrt[n]{n!\plus{}1}}\geq$ $ \frac{n!}{n!\plus{}n}\plus{}\frac{n}{n!\plus{}n}\equal{}1$. But $ n^{n\minus{}1}\geq n!$  Done!

Nice trick with AM-GM ! Personaly, I used Bernoulli's inequality. How did you prove the last one ?

Simply $ n^{n\minus{}1}\equal{}n.n...n$ ($ n\minus{}1$ times) $ \geq$ $ 2.3....n\equal{}n!$


conics
parabola
quadratics
algebra
quadratic formula

Show that the graph of an equation of the form
Cy^2 + Dx + Ey + F = 0, where C does not = zero is:

(a) a parabola if D does not = 0.

(b) a horizontal line if D = 0 and E^2 - 4CF = 0.

(a) If $ D\neq0$, $ x\equal{}\minus{}\frac1{D}(Cy^2\plus{}Ey\plus{}F)$

(b) If $ D\equal{}0$, $ Cy^2\plus{}Ey\plus{}F\equal{}0$, so by using quadratic formula,

\[ y\equal{}\frac{\minus{}E\pm\sqrt{E^2\minus{}4CF}}{2C}\]

Since $ E^2\minus{}4CF\equal{}0$, $ y\equal{}\text{constant}$


calculus
derivative
calculus computations

Hi guys
Can anyone help me out?
Water is flowing into a spherical tank at a constant rate. Let v(t) be the volume of water in the tank and let h(t) be the height of the water in the tank at time t.
Quetion 1: What are the meanings of v'(t) and h'(t)? Are these derivatives positive negative or zero? (Tank empty (t=0))
Question 2: Is v''(t) positive, negative or zero? Explain
Thanks for any help and read the questions!!!

[quote="paladin8"]I don't think this is the right forum... but oh well.

[hide="1."]
Since water is flowing in, the volume and height are both increasing, so v'(t)>0 and h'(t)>0.
[/hide]

[hide="2."]
v''(t) = 0 because v'(t) is not changing (volume is increasing at constant rate).[/hide][/quote]

Thanks a lot for the help man! :)

Wherever this started out, now this is the right forum for it.

A followup question. Consider $h''(t).$

Is $h''(t)$ positive, negative, or zero? Does your answer depend on whether the water surface is above or below the midline of the tank?


A quadrilateral has one vertex on each side of a square of side-length 1. Show that the lengths $a$, $b$, $c$ and $d$ of the sides of the quadrilateral satisfy the inequalities \[ 2\le a^2+b^2+c^2+d^2\le 4. \]

[quote=paladin8][hide]
Let $a^2 = x^2+(1-w)^2$, $b^2 = y^2+(1-x)^2$, $c^2 = z^2+(1-y)^2$, $d^2 = w^2+(1-z)^2$ with $0 \le x,y,z,w \le 1$ (draw a diagram and this will be more obvious; it's just the Pythagorean Theorem). Then

$\sum a^2 = \sum (x^2+(1-x)^2) = 2\sum \left(x-\frac{1}{2}\right)^2+2 = S$.

Since $0 \le \left(x-\frac{1}{2}\right)^2 \le \frac{1}{4}$, we have $2 \le S \le 4$. [/hide][/quote]

Nice solution! I did the same thing, with equalities holding at a,b,c,d the midpoints of the square, and a,b,c,d the vertex’s of it.[quote=JBL]The natural geometric generalizations:  can you find the equivalent bounds for a triangle and a hexagon?[/quote]

The triangle should be 3/4<=S<=3, and hexagon should be 9/2<=S<=6.


let$P \in Z[x]$ and $P(n)>n$ $\forall n\in N$ let $x_{1}=1$ and $x_{n+1}=P(x_{n})$
if there is $n_{N}$ so that N divides $x_{n_{N}}$ prove that $P(x)=x+1$

[quote="jastrzab"]I think I don't understand...how do you define $n_N$ and what is $N$?? Look what will happen if we take $N=1$...[/quote]
                    sorry , N is an abitrary integer number and N > 1

Let's use inductive method to prove: for any positive integer n,P(n)=n+1,then since P(x) is a polynominal,so
P(x)=x+1 for any x.
it's obvious when n=1,
if P(i)=i+1 (i=1,2....k),then when n=k+1,
since P(k+1) is an integer and P(k+1)>k+1,
if P(k+1)>=k+3,let N=P(k+1)-1,then N>k+1, N|(P(k+1)-P(1)),
so P(ak+t)=P(t)(mod N),for any integer "a" and "t",
but none of P(i) (i=1,2....k) is divisible by N (since P(i)=i+1,N>k+1),
so for any integer t, N doesn't divide P(t),a contradiction.
Hence,P(k+1)=k+2 :P


calculus
derivative
function
trigonometry
calculus computations

Hello everybody,
I have a partial derivatives problem that I was unable to prove since I have done it like 4 times and every time it give me "Zero", so if anybody can do it and prove it please send a reply, and thanks in advance. :blush:

Here is the problem:
Show that the function $f(x,y) = xy\frac{x^{2}-y{2}}{x^{2}+y{2}}$ (provided that $f(0,0)=0$), we have $f_{xy}(0,0)=-1$, $f_{yx}(0,0)=1$

You sure you're taking 2 partials, now?  It does work (although it isn't particularly pretty.)  Perhaps you should write up what you've done and we can find where you've gone wrong.

Thanks JBL for your reply! :D
Here is my solution:
$f(x,y)=xy \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$

$f(x,y)=\frac{x^{3}y-y^{3}x}{x^{2}+y^{2}}$

$f(x,y)=(x^{3}y-y^{3}x)(x^{2}+y^{2})^{-1}$

$\frac{\partial{f}(x,y)}{\partial{x}}=(3x^{2}y-y^{3})(x^{2}+y^{2})^{-1}-2x(x^{3}y-y^{3}x)(x^{2}+y^{2})^{-2}$

$\frac{\partial{f}^{2}(x,y)}{\partial{y}\partial{x}}=(3x^{2}-3y^{2})(x^{2}+y^{2})^{-1}-2y(3x^{2}y-y^{3})(x^{2}+y^{2})^{-2}-2x[(x^{3}-3xy^{2})(x^{2}+y^{2})^{-2}-4y(x^{3}y-y^{3}x)(x^{-2}+y^{2})^{-3}]$

and if I am going to put $f_{yx}(0,0)$ then the answer will equal to $=0$ and that is not right since the problem was proving the answer $=1$ and not $=0$

This is my solution, so if there is anything wrong please help me out to answer this problem.
Thank You!

So, I revise my previous statement: those derivatives aren't actually defined: they depend on the direction at which you take your limits as you approach the origin.  In particular, the question wants you to first send $y \to 0$, then send $x \to 0$ (in other words, approach 0 along the $x$-axis), while you've sent it to 0 along the path $y = x$.  (It doesn't say this explicitly, of course, but that does give the answer that it gives.)

In fact, if one approaches along a path of angle $t$, the limit of that second derivative is $\frac{3 \cos 2t-\cos 6t}{2}$.  (This follows by the nice substitution $x = r \cos t, y = r \sin t$ into the expression you've got there.)

First of all, I want to thank both Ramanujan, and JBL for replying to my question, I really appreciate it.

Second, About JBL's solution, I really don't understand anything of his solution. and about Ramanujan's solution after finding $f_{x}(0,y)$ then what do I do? If I get the derivative of it and find $f_{xy}(0,0)$ I will get 1 and not -1 so I must have done something.

PLEASE HELP! :huh:

Thanks Ramanujan for your great help.
 :lol:  :D


If F(X)=[X] and [X] represents the greatest integer less than or equal to X,
evaluate: F(2/3 + F(3/2 - F(-1/2)))

[hide]$f(x)=\lfloor x\rfloor$

$f\left(\frac{2}{3}+f\left(\frac{3}{2}+f\left(-\frac{1}{2}\right)\right)\right)$

$f\left(\frac{2}{3}+f\left(\frac{3}{2}-1\right)\right)$

$f\left(\frac{2}{3}+0\right)$

$\boxed{0}$[/hide]

i_like_pi: that last sign is subtraction, not addition.

[hide]in that case, we have
$f\left(\frac{2}{3}+f\left(\frac{3}{2}-f\left(-\frac{1}{2}\right)\right)\right)$
$f\left(\frac{2}{3}+f\left(\frac{3}{2}+1\right)\right)$
$f\left(\frac{2}{3}+2\right)$
$\boxed{2}$[/hide]

[hide]$f(-\frac{1}{2})=-1$
$f(\frac{3}{2}+1)=2$
$f(\frac{2}{3}+2)= \boxed{2}$

Mentally, you can solve it faster.[/hide]


So can i even understand what the problems on USACO asking for if all the experiences i had in computer science is AP computer science A?  Is it going to be extremely difficult for me?

Can you read and write a file? Excellent. That's all you need to begin with; then train, and train, and train, (with contests liberally interspersed whenever available), and all will be well :). There's a contest from now till Monday, if you want to do some of the training pages over the weekend and then do the contest. Good luck!


This isn't anything close to AP Bio, but i'm wondering what a climatogram is cause i need to create one for a bio project about the african savanna but i don't know what it is? PLZ help!!!

a climatogram is pretty much a graph, chart, statistical representation of weather data from a certain area(s) at certain times. it can include high/low temp., mean temp, avg. barometric pressure, and other weater statistics like precipitation, wind speeds, etc.

plokoon51, an example of a climatograph would look as such:
[img]http://www2.kpr.edu.on.ca/cdciw/biomes/Illinois/abioti1.gif[/img]
Source: [url]http://www2.kpr.edu.on.ca/cdciw/biomes/Illinois/abioti1.gif[/url]

Are you interested in constructing one, for your local area, in an experiment? If so, [url]http://www.ciese.org/curriculum/weatherproj2/en/docs/climatogram.shtml[/url] provides a detailed guide to build a climatograph in Microsoft Excel.


Consider the set E={ f : R -> R>0 | f(x)=$a^{x}$ } a is a constant. Then there is no element in E such that x+w=0. Does it mean that E isn't a vector space?

If I understand the question correctly : it is not a vector space.

However, speaking technically, one can only ask that question when there has been given a definition for the addition....

What about this problem :
Consider the set E={f:R -> R>0 | f(x)+f(y)=f(x)*f(y), x,y are real numbers }
Is E a vector space?
fredbel6 please reply as soon as possible

I think you made a typo there, as I suspect you want to speak about exponential or logarithmic functions.

I would like to emphasize again you have to define operations before you can check whether or not it is vector space.

But in short : it is no vectorspace.

Are you sure that is not a vector space?

$E=\{f: \mathbb{R}\rightarrow\mathbb{R}| f(x)+f(y)=f(x) f(y), x,y \in\mathbb{R}\}$
Latex is your friend :)

Are you sure you don't mean something like :
$E=\{f: \mathbb{R}\rightarrow\mathbb{R}| f(x)+f(y)=f(x y), x,y \in\mathbb{R}\}$

f(x)+f(y)=f(x)*f(y)

The most important is that my professor says it is a vector space. He says I should think in other way :(. I can't understand why 1+1=0 in vector space
{0,1}. Maybe I understood the definition of vector space in wrong way?

Mamat,

this is a vector space...you..just have to check the axioms...
of..it...I think..you'll find them in the lecture notes..posted on the web...

actually...the zero..element..of this space is the function f(x)=1, for all real x..., the additive inverse is 1/f(x)..the other axioms..are checked easily... :)


The strips on the target shown are equal in width. If a randomly-thrown dart lands on the target, what is the probability that it will land within a black strip? Express your answer as a common fraction.

You can't solve this, because there is supposed to be a picture. However, many problems appear to lack these all-important figures...


% MathType!Translator!2!1!LaTeX.tdl!TeX -- LaTeX 2.09 and later!
\[ \begin{array}{l}
 1.If\;a,b,c \ge 0,a \plus{} b \plus{} c \equal{} 1,prove\;that\;(1 \minus{} a^2 )^2  \plus{} (1 \minus{} b^2 )^2  \plus{} (1 \minus{} c^2 )^2  \ge 2 \\
 2.a \ge b \ge c \ge 0,and\;a \plus{} b \plus{} c \equal{} 3,prove\;ab^2  \plus{} bc^2  \plus{} ca^2  \le \frac{{27}}{8} \\
 3.x,y,z \ge 0,x \plus{} y \plus{} z \equal{} \frac{1}{2},find\;the\;\max A,letA \equal{} \frac{{\sqrt x }}{{4x \plus{} 1}} \plus{} \frac{{\sqrt y }}{{4y \plus{} 1}} \plus{} \frac{{\sqrt z }}{{4z \plus{} 1}} \\
 \end{array}
\]
% MathType!End!2!1!

[quote="jacques zhang"]% MathType!Translator!2!1!LaTeX.tdl!TeX -- LaTeX 2.09 and later!
\[
\begin{array}{l} 1.If\;a,b,c \ge 0,a + b + c = 1,prove\;that\;(1 - a^2 )^2 + (1 - b^2 )^2 + (1 - c^2 )^2 \ge 2 \\
[/quote]
$ (1\minus{}a^2)^2\plus{}(1\minus{}b^2)^2\plus{}(1\minus{}c^2)^2 \geq 2$
<=>$ a^4\plus{}b^4\plus{}c^4\plus{}1 \geq 2(a^2\plus{}b^2\plus{}c^2)$
<=>$ a^4\plus{}b^4\plus{}c^4\plus{}(a\plus{}b\plus{}c)^4 \geq 2(a^2\plus{}b^2\plus{}c^2)(a\plus{}b\plus{}c)^2$
<=>$ a^4\plus{}b^4\plus{}c^4\plus{}2(a\plus{}b\plus{}c)^2(ab\plus{}bc\plus{}ca) \geq (a\plus{}b\plus{}c)^2(a^2\plus{}b^2\plus{}c^2)$
<=>$ a^4\plus{}b^4\plus{}c^4\plus{}4(ab\plus{}bc\plus{}ca)^2 \geq (a^2\plus{}b^2\plus{}c^2)^2$
This inequality is true!
Done!

[hide]problem number 2 , i think it was posted several times before  :)

problem 3
expand the obvious inequality : $ \frac {1}{1800}\frac {(6a \minus{} \sqrt {6})^2(12a^2 \plus{} 4a\sqrt {6} \plus{} 27)\sqrt {6}}{4a^2 \plus{} 1} \geq 0$(with $ a \equal{} \sqrt {x}$) we get $ \frac {a}{4a^2 \plus{} 1} \leq \frac {\sqrt {6}}{10} \plus{} \frac {1}{100\sqrt {6}}(36a^2 \minus{} 6)$
setting up 2 other similar ineq we have
$ \sum \frac {a}{4a^2 \plus{} 1} \leq \sum \frac {3\sqrt {6}}{10} \plus{} \sum\frac {1}{100\sqrt {6}}(36a^2 \minus{} 6) \equal{} {3\sqrt {6}}{10}$[/hide]

% MathType!Translator!2!1!LaTeX.tdl!TeX -- LaTeX 2.09 and later!
\[ problem\;3: \;the\;answer\;is\;\frac{3}{5}\sqrt {\frac{3}{2}}  \equal{} \frac{3}{{10}}\sqrt 6
\]
% MathType!End!2!1!
thanks :) i have got it


AMC
USA(J)MO
USAMO
algorithm
AMC 10
AMC 12
AMC 12 A

Again this year, participants in this section of the Community Forum are debating the merits of the published process for USAMO selection.  Last year, in response to same debate, the staff at the American Mathematics Competitions office conducted an internal study of 9 proposed alternative methods for USAMO selection.  I have put that internal study up on our website.

I quote from the conclusions of that study: "Compared to the other proposed USAMO selection methods, the current method does not appear to be any more or less “fair” or indicative of USAMO success"  and "It still appears that no other selection method is significantly better than the current method."

In response to a couple of other comments that have appeared in this Forum:
[quote]For some substantial proof, take a look at the Wikipedia site for USAMO.[/quote]
The official statements about policies and procedures for the AMC contests are in the current year Teachers Manuals which are repeated on our website.   The AMC staff has had no part in the writing or editing of the Wikipedia article.

[quote]Does anyone know if the MAA officials even read some of the posts here [/quote]
Yes.

Steve Dunbar
MAA Director, American Mathematics Competitions

With all due respect, I have a few questions:

I just read the study that has been put up on the AMC website. The study seems to correlate several methods with USAMO success, and the study seems to have determined that "Compared to the other proposed USAMO selection methods, the current method does not appear to be any more or less “fair” or indicative of USAMO success." This was based purely on the correlation between the index created and USAMO success.

First of all, there is a simple algorithm that could be implemented that was not mentioned in the paper (at least, I didn't see it). And in fact, this algorithm is being utilized by the AMC directors in their selection process, only not for the selection of the underclassmen. Currently, when the first "wave" of USAMO qualifiers are being selected, the MAA states that: "To adjust for variations in contest difficulty, the number of students selected from A & B contests will be proportional to the number of students who took the (A & B) Contests." Can't this be applied to the underclassmen who take the AMC10 versus the AMC12? It is, in essence, the same idea, to "adjust for variations in contest difficulty". And in actuality, the "variations in contest difficulty" between the AMC12 and the AMC10 is far greater than the variations between the AMC12A and the AMC12B.

Also, how significant is the calculation of the correlation between various indicies and USAMO success? Would a new, better algorithm really have a noticeable impact on this correlation constant? The study mentions the "‘any given day’ effect". No algorithm can be created, no matter how complicated, that will perfectly (or even close to perfectly) model the performance of the student on the USAMO. Instead of looking for the correlation constant to change, we can look at the pure mechanics of what is happening during the selection process.

The plain truth is, an underclassman who has a 132 AMC12 / 7 AIME is (under the current system) equivalent to an underclassman who has a 132 AMC10 / 7 AIME. Clearly, this is unfair. The difference between a 132 AMC12 and a 132 AMC10 is great, and in some years (like this year), even greater. So it is possible that some students will be unfairly accepted before others.

Some may question whether this really matters. Look at the correlation constant. It doesn't change much. Think of all the other students who got into USAMO. Comparatively, the number of students unfairly deprived of USAMO participation is negligible. Changing this algorithm will not matter in the long run, the USAMO winners will not change, the IMO team will not change, the IMO results will certainly not change.

But I don't think this is a matter of whether it matters in the long run. I just think it's unfair to those who miss the cutoff for the USAMO, not because they fall behind in mathematical ability, but because they decided to (and, in many other cases, actually had no choice but to) take the AMC12 over the AMC10. Undoubtedly these cases occur. And whether these cases are one in 100 or one in 1000, the point is that this isn't right.

The study shouldn't have based its conclusions off of correlation constants. Correlation constants do not show the individual students who unfairly miss qualification.

And as a final note, as someone else pointed out in another thread, this obvious favoring of the AMC10 over the AMC12 for underclassmen actually encourages underclassmen who have the opportunity to challenge themselves to take the AMC10. This is unfortunate, as they should be taking the test that challenges them most, not the test that they think will catapult them into the USAMO.

Please don't take this post as disrespect to the MAA and its current selection process. I am simply pointing out a flaw in the process that, in my opinion, should be fixed. I would be very indebted to you if you considered my request. Thank you!

If you made lots of stupid errors on this AIME, then under similar circumstances you would have an equally high chance of making just as many stupid errors on the AIME 2.

Of course you know which part of the report I read first!

(This is all dealing with 2007 data.)

[quote="The report"]2. There were 42 high AMC10 scorers who were picked in the initial selection.  These 42 almost exactly replace the 34 “higher than floors”.  Those 34 all had low to even marginal AMC12 scores, so had low M.M. indices.  (The student with the 10 on the
AIME had a 99 on AMC12A for a M.M. index of 201.5)[/quote]
That's that part that I should have anticipated and didn't. All right, suppose we have a 10th grader with an AMC 10 of 138 and an AIME of 11. We all agree that that person should be invited to the USAMO - every actual or proposed method agrees on that much. The question is at what stage of the process he should be selected and whom he should displace. Under the current system, his selection is postponed to the second round; once you have him in the second round, his huge index will drive the required second-cut index higher. If you include him in the first round, he displaces some 11th or 12th grader who was a marginal qualifier under the current system, causing the first-cut index to creep upwards.

My take on this: someone whose selection is obvious should be in the first cut. But since there are such students, the best thing to do is to accommodate them by increasing the size of the first cut - say to 360 or 370 instead of 330. That would lower the first-cut index back down to roughly what it would have been under the current system and correspondingly raise the index for the second cut. (But with the stars removed from the pool and the AMC 10 scores devalued, the second cut index would still be down below 180 in 2007 terms.) And that 99+10=201.5 would be back in with the first-cut index back to 200 or below.

[quote="The report"]It is possible that with a further step of making the first selection based only upon AMC12 takers would catch the “higher than floors” omitted and select the ‘proper’ students at the floor to encourage AMC12 taking over AMC10 for USAMO hopefuls.[/quote]
Yes, that does partly address the perverse incentive question - but I'm not a fan. I don't want the overwhelmingly-qualified mucking up the standards for the second cut.

[quote="The report"]A question for debate remains whether a high AMC12 score and a
4 AIME will do better on the USAMO than a “middling” AMC10 score and a 6 AIME.  By
the M.M. a 132 AMC12 + 4 AIME (174.5 M.M.Index) got in while a 129 AMC10 + 6
AIME (172.0 M.M.Index) did not.[/quote]

Of course, if the second cut pulled in 30 or 40 fewer people, they might both be on the outside. I agree, a 4 AIME is pretty low - but a 132 AMC 12 is a rare high score. A 6 AIME is better, but a 129 AMC 10 is only barely an AIME qualifying score. If you want to select only on AIME, then do so. If you want the AMC to count at all, then you are playing the two rounds off against each other. Throw someone with a 144 AMC 10 and 5 AIME into that mix (I knew someone like that in that year). That's another M.M. index in the mid-170's.

[quote="The report"]So the correlation seems to be slightly better for the M.M., however,  21 of the omitted official list students scored 10 or greater including a 16 and an 18.  Most  of those students had taken the ‘harder’ AIME I.[/quote]
If there's a measurable discrepancy between the AIME I and AIME II, I think that needs to be corrected for in some fashion before applying any other method. I think this problem was specific to 2007; having discrepant AIME I and AIME II is a bad thing for any system.

There are many students who never get invited to take the USAMO who would do well on it given the chance - and there always have been. Olympiad math is somewhat different than the AMC or AIME. The expansion from 200 to 500 was a good thing in that it gave some such students a chance. That came at a cost, of course - not just the cost of sponsorship, but that expansion also dragged in some students who don't even want to take the USAMO, and it greatly increased the number of schools asked to assume the burden of administering the USAMO (a burden which some schools have refused).

[quote="AMCDirector"]
[quote]For some substantial proof, take a look at the Wikipedia site for USAMO.[/quote]
The official statements about policies and procedures for the AMC contests are in the current year Teachers Manuals which are repeated on our website.   The AMC staff has had no part in the writing or editing of the Wikipedia article.
[/quote]

However, the numbers are still correct, unless they were fabricated and put up on the Wikipedia site by some strange person for some strange reason.

As a variant, I like the idea of having not two cuts but 4: a first cut for everyone (unified index, or "M.M. index," of course); then a second cut for 11th grade or younger; then a third cut for 10th grade or younger, then a fourth cut for specifically 9th graders; then state winners. The exact numbers to put in each of these cuts - haven't worked that out yet.

Why give an extra chance to 11th graders? Because they have a greater chance of being invited to MOP. Why give an extra chance to 10th graders? Just the reasons why that is already in place. Why give yet one more chance to 9th graders? Two reasons; the first is a wider cast of the net for Red MOP; the second is so that a 9th grader can have a chance to experience the USAMO, do very badly on it, and still have time to grow into a better performance.

[quote="wuyonglin2004"]I think the rule should be changed, but it shouldn't be changed this year as many students have already taken AMC 10 because of the policy, then it will be unfair to them if the rule is changed.[/quote]

It wouldn't be unfair to them. It would simply no longer be unfairly advantageous for them.

[quote="thing1"]I have a question why can't the AMC board allow people to take both AIME's I [b]really[/b] screwed my AIME score up with those arithmetic errors.[/quote]

I have learned that checking mistakes and working carefully is just as important as knowledge and intelligence. But still, I no this sounds crazy, what if they had a trick to check ur answer, like (sum of digits of answer)^(answer mod7) mod 13 or something like that thats too hard to use as the problem?

[quote="crazypianist1116"]Question: Why do we have a cut for state winners who were not originally accepted? Have there been any recorded cases in which state winners accepted by this process have done surprisingly well at USAMO?[/quote]

I am not sure about recorded cases, but the purpose of this is so that every state/region is represented in the USAMO. You may judge if this is fair or not for yourself.

[quote]Have there been any recorded cases in which state winners accepted by this process have done surprisingly well at USAMO? [/quote]
In 1998, the Mississippi state winner, who would not otherwise have qualified for the USAMO, was invited to MOP. (Then he moved to California and qualified more conventionally the next year.)

There are almost certainly people who would do surprisingly well at USAMO who are nowhere close to qualifying for it.

I am happy that the AMC takes USAMO qualification so seriously!  I think that with both AMC 10 and AMC 12 tests, there is probably no perfect selection mechanism, and whatever method is used, there will probably be people who were close to qualifying but didn't---perhaps wrongly---and they will be unhappy about that.

Thanks to the AMC for making such a thorough investigation.  I find it interesting that percentiles tend to correlate less with USAMO score than actual index.  Perhaps this means that in a given year, the difficulty of the AMC tests is more consistant than the ability of the audiences taking them!

[quote="Kent Merryfield"]In 1997, the Mississippi state winner...[/quote]1998.

On the right scaling for the unified-index system and multiple cuts:
This was used in the late 1990s, and 1999 data is on the website. The standards then were 120 in the first cut, 10 $ \le$ 11th graders, 10 $ \le$ 10th graders, 10 $ \le$ 9th graders, and state winners. With a few ties and 20 making it in the last step, this resulted in 171 total qualifiers. Of the 151 before state qualifiers, 7 were 8th grade or younger, 19 9th grade, 29 10th grade, 42 11th grade, and 54 12th grade. The cutoff levels were 201, 193, 185, 173.
Tripling all of those numbers for the current target of ~500 is a decent place to start.

From the USAMO Teachers' Manual:
[quote]Please fax at least one official USAMO answer sheet with student number for each problem, even if the answer sheet is blank because the student did not attempt a solution. This assures that we have received all problems from all students.[/quote]

[quote="bookaholic"]Is it just me, or does the breakdown of cutoffs make the unlikely situation we were discussing earlier in this thread scarily probable? Suppose someone has a 132 AMC-12, 144 AMC-10, and 7 AIME. If only their AMC-10 score is considered, wouldn't they miss USAMO despite having a high enough AMC-12 index? Was this factored in?[/quote]

Yes. They take whatever score will make you qualify.

It sounds to me like when they're computing the cutoffs, they consider only each student's higher score.  But I'm sure if that happened to someone, they could just contact the AMC and be added to the list.

Team Selection Test. It's the final round the 12(+) winners take to determine who the 6-person IMO team is.

By getting an index of 192.5 (112.5/8), I missed the cutoff by one AIME question this year. (I missed #1, 807) And because I'm in the 11th grade, even though I got an AIME score of 8 or more, and qualified through the AMC 12, I missed the cutoff by being one grade too high (I skipped a grade).

Last year I had an index of 197, and the cutoff was 202.5 for 10th graders or lower. One AMC question away. (I missed #2 on the AMC 10A that year)  I also got an AIME score of 8 that year, and would have certainly qualified through that selection process they newly implemented (the floor was only 6).  I'm not trying to whine, but it helps if I share this with people.

[quote="buzzer11"]Haha, I got a 126/7 as a 10th grader this year (taking the amc12), and I also missed AIME #1. :([/quote]
darn so close

[quote="FantasyLover"]Sigh...

I also didn't make USAMO because I missed #1 on AIME...

I had 126/8=206 as an 8th grader and if I had gotten #1 right, then I would have with 216 index...[/quote]
thats too bad u thought u were going to make it :(

[quote="buzzer11"]I'm guessing you took the AMC10?[/quote]

Yes.

I took AMC 10A and 10B, both resulting in 126...


Here are some identities for which I can't find their solution! :(

1) prod(k=0,n) (2cos(x/2^k)-1)=(2cosx+1)/(2cos(x/2^n)+1);
2) prod(k=1,n) (2cos2^(k-1)x-1)=(2cos2^nx+1)/(2cosx+1)
3) prod(k=1,n) (cosx/2^k+cos(y/2^k))=(cosx-cosy)/(2^n((cos(x/2^n)-cos(y/2^n)))
4) prod(k=0,n) cos2^kx= sin2^(n+1)x/(2^(n+1)sinx);
5) prod(k=1,n) (1+1/cos2^(k-1)x)=tan2^(n-1)x/tan(x/2).

thank you very much! :)

for the 1st one use that (2cos2a+1)(2cos2a-1)=4(1+cos2a)/2-1=2cos2a+1 etc.

cheers! :D :D

for the 2nd one it's too simple use a)! :)

cheers! :D :D

for the 3rd one use the fact that 2(cosa-cosb)(cosa+cosb)=2(cos^2a-cos^b)=1+cos2a-1-cos2b=cos2a-cos2b. and you replace a=x/2^k and b=y/2^k and you multiply all the relations for k=1 to n and you get the identity...

cheers! :D :D

for the 4th one you just use that cosa=sin2a/(2sina) and you replace a=2^kx and you multiply all the relations for k=1 to n...

cheers! :D :D

for the 5th one prod(k=1,n) (1+1/cos2^(k-1)x)=prod(k=1,n) (2cos^2 2^(k-2)x)/cos2^(k-1)x=
prod(k=1,n) tan2^(k-1)x/tan2^(k-2)x=tan2^(n-1)x/tan(x/2)

all problems done!! :D :D :D :D  hope you understood! :D :D they are simple but the need just a little bit of attention from you! :D :D


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

What is the closure of $\{M^2 : M \in GL_n(\mathbb{R})\}$ in $GL_n(\mathbb{R})$ ? and if one replaces $GL_n$ by $M_n$ ? Are there a nice description of these sets ?

Let E={M^2 / M is in GLn(R)}=exp(Mn(R)) and F be the set of real matrices M s.t. the Jordan's decomposition of M has the following property (P): if there exists a Jordan's block of odd dimension with a negative (<0) eigenvalue then this block occurs an even number of times.
   Let sp() denotes the spectrum of a matrix;let c(E) denotes the closure of E in Mn(R); the changes are easy for the closure in GLn(R).
    We show that F is included in  c(E) ; perhaps the equality stands.
    1) E is included in F  because if the Jordan's decomposition of a real matrix contains a Jordan's block with an eigenvalue i.t (t real) then it contains its conjugate as another Jordan's block.
 Remark: if A is in E or c(E) then their negative eigenvalues have even multiplicity.
    2) Let A be in F; A=B+B1+B2+B3 as a direct sum where
sp(B),sp(B1),sp(B2),sp(B3) are included in R*-,R*+,{0},C\R.
B1 and B3 are in E. B2=lim(B2+(1/n).id); B2+(1/n).id is in E as B1; thus B2 is in c(E). The problem is in B.
    3) We may suppose that B is a direct sum of Jordan's matrices.Let J a such matrix:J(i,i)=a<0, J(i,i+1)=1, the others 0.
case 1: the dimension k of J is even; let Jx the matrix s.t. Jx(i,j)=J(i,j) except Jx(n,1)=x. Its characteristic polynomial
(z-a)^k-x  has no real root if we choose a negative x.
lim Jx=J when x->0- and Jx is in E (as B3 did); thus J is in c(E).
Remark: for example [-1,1,0,-1] (line by line) is in c(E)\E.
case 2: k is odd; then (P) implies that there exists a second block like J (a twin). J+J, as a direct sum, is in E because J+J=M^2 where we may choose M as follows:
M=[0,U,V,0] (in blocks line by line) and the (k,k) matrices U,V  are polynomials in the nilpotent J-a.id.       CQFD.

     Problem: can you find a matrix in c(E)-F?????


logarithms
function
LaTeX
superior algebra
superior algebra theorems

Hi guys. Please help me. I'm stuck.

Determine whether $ \phi$ is an isophormism.
Let $ <R,\plus{}>$ map with $ <R, \bullet>$ where $ \phi(r) \equal{} 0.5^r$. Im just having trouble with the onto part. Sorry about the bullet.  :D

Okay so to start off the we want to find some $ x \in \mathbb{R}$ such that $ \phi(x)=y$ for all $ y \in \mathbb{R}$.
The [i]isomorphism[/i] $ \phi : <\mathbb{R}_1,+> \rightarrow <\mathbb{R}_2, \times>$ sends elements according to the rule $ \phi(x) = (\frac{1}{2})^x$

Lets take some $ y \in \mathbb{R}_2$. (I am going to subscript it so it is easily seen which group I am referring to when). Lets refer back to logarithm laws.
We have $ a^{\log_a x} = x$ So can you think of what element of ${ \mathbb{R}_1}$ we're going to want to take now, to be sent to some given $ y \in \mathbb{R}_2$ ??

[quote="Fermat's Little Turtle"]Okay so to start off the we want to find some $ x \in \mathbb{R}$ such that $ \phi(x) = y$ for all $ y \in \mathbb{R}$.
The [i]isomorphism[/i] $ \phi : < \mathbb{R}_1,+ > \rightarrow < \mathbb{R}_2, \times >$ sends elements according to the rule $ \phi(x) = (\frac {1}{2})^x$

Lets take some $ y \in \mathbb{R}_2$. (I am going to subscript it so it is easily seen which group I am referring to when). Lets refer back to logarithm laws.
We have $ a^{\log_a x} = x$ So can you think of what element of ${ \mathbb{R}_1}$ we're going to want to take now, to be sent to some given $ y \in \mathbb{R}_2$ ??[/quote]

I'm a little confused, but anyway, please tell me, if my solution is correct. I didn't follow yours.

If I let $ y \in \mathbb{R_2}$, then $ y = (\frac{1}{2})^x$. I think this is just an inverse function, right?

So get the logarithm of both sides. We have $ \log(y) = x \frac{1}{2}$ which implies that $ x = \frac{\log(y)}{\log(\frac{1}{2})}$

It is just the inverse function =) Yea.
By the way I didn't post a solution, just an idea of how to get it. Also you can write $ x \equal{} \minus{}\frac{\log(y)}{log(2)}$ (I find it looks better :P)

First, some LaTeX: \langle and \rangle give $ \langle$ and $ \rangle$, much nicer than < and >.  \cdot is a multiplication dot, \times is the multiplication X.  Be careful with fonts: you wrote \mathbb{R_2}, and this came out looking very strange -- you should have used \mathbb{R}_2.

The two algebraic objects are clearly not isomorphic because $ \langle {\bf R}, \plus{}\rangle$ is a group but 0 has no inverse in $ \langle {\bf R}, \times\rangle$.  The map $ x \mapsto \frac{1}{2^x}$ is a bijection between the real numbers and the positive real numbers.  All parts of the problem are easy in this case -- the inverse function of exponentiation is indeed the logarithm, etc.


limit
real analysis
real analysis unsolved

$u_{n}>0$ for all n positive integer

Find the nature $\sum\frac{\ln(2+e^{|u_{n+1}-u_{n}|})}{u_{n}}$

Perhaps you mean the numerator to be
[quote]$\ln(2+e^{|u_{n+1}-u_{n}|})$[/quote]

It is divergent.
Proof:
Let $a_{n}={\ln(2+e^{|u_{n+1}-u_{n}|})}{u_{n}}$
(1),$\lim_{n \to+\infty}u_{n}\neq+\infty$,then we have:
$\lim_{n \to+\infty}\frac{\ln(2+e^{|u_{n+1}-u_{n}|})}{u_{n}}\neq 0$
Thus $\sum a_{n}$ is divergent.
(2)$\lim_{n \to+\infty}u_{n}=+\infty$
we have for every $N$:
$a_{N+1}+a_{N+2}+\cdots+a_{n}\ge \frac{u_{n}-u_{N+1}}{u_{n}}$
let $n \to+\infty$
We will get:
$\frac{u_{n}-u_{N+1}}{u_{n}}\ge \frac{1}{2}$
So the series is also divergent. :)


Please see the attachment for the picture.
The problem is: A diagonal is drawn in rectangle ACEG, as shown. Segments BF and HD are drawn parallel to sides CE and GE, respectively, with segments AE, BF and HD intersecting at point I. the area of rectangle BCDI is 36, What is the area of rectangle FGHI? (How to get the answer also please)

Let $ BI \equal{} a$, $ ID \equal{} b$, $ BI \equal{} x$, and $ IF \equal{} y$.

Since triangles ABI and IFE are similar, we have $ \frac{a}{x}\equal{}\frac{b}{y}\implies{ay \equal{} bx}$. We are already given that $ ay \equal{} 36$ (area of BCDI) so $ bx$ (area of FGHI) is also $ \boxed{36}$

If we let $ BI \equal{} CD \equal{} AH \equal{} x$, then $ BC \equal{} DI \equal{} EF \equal{} \frac {36}x$ (do you see why?).

Now, it's similar triangle bash. Keep in mind that we want to find $ yz$ (the area of rectangle $ FGIH$).

Since $ \triangle EFI\sim \triangle IHA$, we have $ \frac {\frac {36}x}y \equal{} \frac zx$, so cross multiplying,

$ [FGIH] \equal{} yz \equal{} \frac {36}x\cdot x \equal{} \boxed{36}$.

Edit: Beaten, but diagram attached. :P

Thank you so much, your solutions were better than mine


Buenas tardes,

Como se ha hecho en anteriores ocasines ...
aqui propongo una encuesta ..
haremos uso del siguiente esquema:
[b]Esquema: PAIS {Ptos. en IMO 2006}[/b]

[b]PD: Se ha considerado los resultados del ultimo año
para hacer la lista[/b]

Yo digo que va a ser colombia pues tenemos 5 posibles personas repitiendo IMO. El año pasado era la primera IMO de todos.

JAS

Hola jasamper88,

SERIA ESTUPENDO
que dees tu voto ;)

saludos,

ok

he agregado a COLOMBIA,
 disculpen si no aparecieran los otros paises,
pero he considerado a los 7 primeros paises latinoamericanos
segun el listado por paises de la IMO 2006.

SALUDOS,

[quote="R.G.A.M."][color=blue]Puede que México dé nota nuevamente este año, ya que sinproblemas fue el mejor equipo íberoamericano tanto de la I.M.O. como de la O.I.M..

Me intriga saber por qué no incluyeron a Paraguay, sabiendo que tuvo un muy alto rendimiento el año pasado (considerando que asistimos cuatro participantes). Comento, de antemano, que muy probablemente sólo cuatro jóvenes conformen la delegación paraguaya este año, nuevamente.[/color][/quote]

Buenos dias R.G.A.M.,
aunque es cierto que el ratio ptos/participantes de PARAGUAY es alto,
he explicado que se han considerado la puntuacion global de equipo..

Saludos,

Se hace notar la mayor presencia peruana en el foro  :P

[color=blue]No dudo que el equipo peruano se va a destacar, mas considero que necesitaría contar con un mínimo de tres candidatos a medalla de plata para ser el mejor equipo íberoamericano en la I.M.O..[/color]

[color=blue]Todos los años ganan los peruanos en esta encuesta.
Sospecho que la superpoblación de ellos en este foro es la causa.

Es casi definitivo que Paraguay estará fuera de esta competencia para el próximo año, ya que vamos cuatro participantes, nuevamente.[/color]

Bueno aqui estan los resultados de los latinos:

1) Brazil (106 puntos) (2 platas, 3 bronces)
2) Colombia (93 puntos) (1 plata, 3 bronces)
3) Peru (91 puntos)  (1 plata, 2 bronces)
4) Mexico (86 puntos) (4 bronces)
5) Argentina (75 puntos) (1 plata, 1 bronce)

Este año hizo falta un oro latino.

y Chile ???

ajaj era broma, q mal le fue al equipo chileno...felicidades a uds.

saludos


Let $ABC$ be a triangle.  I hope this isn't too brutal: prove that
\[ \sin^2 A (1+\cot B \cot C) + \sin^2 B (1+\cot C \cot A) + \sin^2 C (1+\cot A \cot B) = 3.  \]

ahahah.. thats no hint man.. thats giving the entire thing away^^  :P

nice problem..


The n integers 1,2,...n are arranged around a circle.Let A denote
the number of such arrangements in which no two adjacent integers are consecutive in clockwise order ( n and 1 are considered to be consecutive) .

would anybody help me to find the expression for A,please?
Thank you so much!!

[quote="Elemennop"]$A=n!\sum_{i=0}^{n}\frac{(-1)^{i}}{i!}$[/quote]

It works for all integer $\,n$? Let $n=3$ for example.

(1, 3, 2) 1 and 2 aren't adjacent ?

132 would work since reading clockwise they aren't consecutive...however I do believe the formula doens't work since it's circular..

[quote="kathy"]( n and 1 are considered to be consecutive) .[/quote]

1 and 2 in (1, 3, 2) are consecutive and adjacent if you take it on a circle.

Reading clockwise - 1 and 3 are not consecute, 3(n) and 2 are not consecutive, 2 and 1 are not consecutive (since 2=/=n).


function
calculus
algebra solved
algebra

Without using calculus determine the least value of the function

f(x)=(x+a+b)(x+a-b)(x-a+b)(x-a-b) where a,b are real constant


Let  $ P\in\mathbb{IR}[X],$ $ degP\equal{}n\minus{}1$  such that there exist $ r>0$ with $ P(k)\equal{}r^{k}$ for every $ k\in\{1,2,...,n\}.$

Calculate $ P(n\plus{}1)$


algorithm
combinatorics proposed
combinatorics

I like this game (I hope I didn't already posted here) :

Assume you have an infinite string of unit squares numbered 1,2,3,... from left to right. Initially, there is a coin on the square numbered 2. In turn, Alice (first) and Bob move the coin to the right by any number of squares they are willing to, but at least one and no more than 2003. The first who places the coin in a square whose number is not a prime looses the game.

Prove that the game will eventually end and determine which player has a winning strategy.

Pierre.

[quote="pbornsztein"]Prove that the game will eventually end[/quote][hide="it will end but it might take a while ... how old are Alice and Bob?"]I can think that it will end in at most 2002! steps :D (the first 2003 consecutive integers which are not prime that come to my mind now are 2003!+2, 2003!+3, ... , 2003!+2004 :P).[/hide]

It's funny : I was knowing two kind of solutions for this problem.
First one by working backward from the first sequence of at least 2005 consecutive composite numbers.
Second by strategy stealing : suppose that the second player has a winning strategy. Then, the first one plays 3 and since 2006 is composite, the first player can use (steal) the strategy of the second and so has himself a winning strategy. A contradiction. Thus, the second has no winning strategy, so that the first one has one (since it is a complete information finite game. I like the startegy stealing argumentation).

Myth's approach seems to be a patchwork from these two  :)

Pierre.

[quote="Myth"]Using backtracking algorithm we can mark all natural numbers with '+' and '-', where '+' is a winning position for player, who obtained this number after his turn, '-' otherwise.
We will prove that there is '+' in set 3, 4, 5, ..., 2005 (clearly, first player wins in this case).
Suppose contrary, i.e. all numbers 3, 4, 5, ..., 2005 are losing. Then first player moves coin to number 3, and then second player can move it to 4, 5, ..., 2005 or 2006, but 2006 is composite, so possible moves for second player are 3, 4, ..., 2005, but all these numbers are losing!!! Contradiction.[/quote]Wow! I love this CS-enabled solution :)

[quote="pbornsztein"]Myth's approach seems to be a patchwork from these two  :) [/quote]
I have looked to my dictionary, and I understood that I don't like word "patchwork" :)  :D  :D  :P


search
conics
parabola
geometry
Pythagorean Theorem

I have 3 old MC problems.
On 1995 #29, I managed to guess it right (go powers of 2), but yeah.

[hide]1.Use the fact that $ \triangle TOL\sim \triangle CNO$.

2. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1129280303&t=195357[/url]

3. Flip the parabola over in order to exterminate the negative values.
Let Point A be (X,Y)
Substitute $ y\equal{}\frac{x^2}{2}$

Use the fact that altitude can be expressed in two different ways.
$ x\sqrt{3}\equal{} \equal{}\frac{x^2}{2}$
Solve to get $ x\equal{}2\sqrt{3}$
Times $ x$ by two to get the answer.[/hide]

[hide="no.19"]#19. $ \angle TOL \plus{} \angle OLN \plus{} \angle NOC \equal{} 180$degrees. Since $ \angle LON$ has measure 90 degrees, and $ \angle OLT \plus{} \angle TOL \equal{} 90$ while

$ \angle CNO \plus{} \angle NOC \equal{} 90$(right triangles)...

$ \angle OLT \equal{} \angle CON$, which implies that $ \triangle OLT$ is similar to $ \triangle CON$. Thus $ OC$ is $ 7$.

With that and the Pythagorean Theorem,

$ ON \equal{} \sqrt {1^2 \plus{} 3^2} \equal{} \sqrt {10}$ and $ OL \equal{} 7\sqrt {10}$.

Thus the area is $ \sqrt {10}*7\sqrt {10} \equal{} 70.$[/hide]

[hide="no.29"]It seems like there's a pattern... the rows i've found all are exponents of 2...but it baffles me why;please post solution  :blush:(my answer is 4) [/hide]

[hide="no.3"]In $ A$, $ x \equal{} \frac {y}{\sqrt {3}} \equal{} \frac {y\sqrt {3}}{3}$ since the triangle formed with A, the origin and the the intersection of AB and the y-axis is a 30-60-90 right triangle.

Substituting the equation...

$ y \equal{} \left( \minus{} \frac {1}{2}\right){\left(\frac {y\sqrt {3}}{3}\right)}^2$
$ y \equal{} \left( \minus{} \frac {1}{2}\right)\left(\frac {y^2}{3}\right)$
$ y^2 \plus{} 6y \equal{} 0$
$ y \equal{} (0, \minus{} 6)$

Thus the length of the triangle is $ 28\frac {6}{\sqrt {3}} \equal{} 4\sqrt {3}$[/hide]


[quote="anirudh"][quote="nebula42"]5 copiers can copy 34,000 sheets in 6 hours. At the same rate, how many hours will take 15 copiers to copy 408,000 sheets?

(Slightly More Difficult) Generalization:
If $a$ copiers can copy $b$ sheets in $c$ hours, and $d$ copiers can copy $e$ sheets in $f$ hours, find an equation relating $a,b,c,d,e,$ and $f$.[/quote]

[hide]
for the first one, is it 24 hours?[/hide][/quote]
Yes, that's correct  :) , but could you please provide a solution?  :wink:
Also, now that you've solved the first problem, try to solve the generalization.

[quote="nebula42"]5 copiers can copy 34,000 sheets in 6 hours. At the same rate, how many hours will take 15 copiers to copy 408,000 sheets?

(Slightly More Difficult) Generalization:
If $a$ copiers can copy $b$ sheets in $c$ hours, and $d$ copiers can copy $e$ sheets in $f$ hours, find an equation relating $a,b,c,d,e,$ and $f$.[/quote]

[hide]so 15 copy 102000 in 6 hours. thats 24 hours then[/hide]

[quote="nebula42"]5 copiers can copy 34,000 sheets in 6 hours. At the same rate, how many hours will take 15 copiers to copy 408,000 sheets?

(Slightly More Difficult) Generalization:
If $a$ copiers can copy $b$ sheets in $c$ hours, and $d$ copiers can copy $e$ sheets in $f$ hours, find an equation relating $a,b,c,d,e,$ and $f$.[/quote]
[hide="Number 1"]$\frac{30}{34000}=\frac{15x}{408000}$

$12240000=510000x$

$x=24$[/hide]

[hide="Generalization"]$\frac{ac}{b}=\frac{df}{e}$[/hide]

[hide]b/ac=e/df. Since in c hours a can print b sheets. Every printer prints b/a sheets in a given c hours. Thus b/ac represents the amount of sheets a printer prints in an hour. Its the same for the other printers and sheets.[/hide]


There are 12 pieces of paper in a bag. There are 4 blue pieces, 4 red pieces, and 4 black pieces. There are a number of white dots on each piece of paper. The first blue piece has one white dot, the second blue piece has two white dots, the third has three, and the fourth has four. This is the same with the other colors of paper.
5 randomly chosen pieces of paper are drawn. What is the chance that three pieces of paper have the same number white dots, and the other 2 have the same number of dots too (but they are a different amount than the other three)?

[quote]If home is where the heart is and my heart is in this universe, what are you doing in my house?[/quote]

I believe without, but then the probability is a bit on the small side...

i doubt this is right but:

[hide]first one you choose doesn't matter so it would be 12/12

next 1/11 and 1/10. next one doesn't matter since there are only 3 letters with same white dots so it's 9/9. and 2/8.

1/11*1/10*1/4=1/440?[/hide]

Leeyuhang -- you have (possibly among other problems) the same problem you had before -- how many choices do you really have at each opportunity?  1/11 isn't going to do it.

Another thing you should always be careful of -- check out when you are having order be important and when you are not.  I can take what you started with a find the answer, but I recommend that you start over and go through it, very carefully.  Another thing you might want to try is to count the number of good hands and divide by the number of total hands -- that helps avoid the first problem and makes the second problem easier to keep track of.  Probability is hard -- you should always make sure you can justify everything you do.

Tell me this -- what kind of order did you take into account?  That is, much order did you restrict you answer to, and is that the appropriate amount?

are you saying in problems like these you have to put into consideration how many ways you're numbers can be arranged?

Yup, for example there's 6 ways to pull out 3 balls if order counts, 123, 132, 213, 231, 312, and 321. Usually the answer is n!, do you see why?

so for the pulling 2 out the same would it be multiplyed by 3 because there are 3 combinations?

I'm not quite sure what your question was, but let me tell you what the ordering you did was.  You first took the three with one number, then the two with the other number.  So your two groups were ordered.  However, within each group, you did not order -- your calculations accounted for blue-red-black, bl-bk-rd, rd-bl-bk, rd-bk-bl, bk-rd-bl, and bk-bl-rd, as well as all the possible combinations for the pair.  So you partially ordered and partially didn't order.  But the question didn't want you to do it like that, so you came up with a different answer.
Let me show you how I did it, and then show you how to get from what you did to the final answer.

Let's compute the probability like this:  first, we'll calculate the number of "good hands," then the number of total hands, then divide.  Any good hand must contain 3 of one number and 2 of another.
We have 4 choices for the number of dots which will be our three-of-a-kind.  After making that choice, we have to take the paper with the appropriate number of dots from each color, and there is only one way to do this.  Thus, we get 4 things here.
Now, we have 3 choices left for the number of dots which will be our pair.  After choosing that, we have to pick two pieces of paper from 3 available colors -- there are 3 ways to do that.  This gives us 9 ways to pick out pair.
This means we have 4*9 = 36 total ways to pick both the three-of-a-kind and the pair.  Now, there are going to be exactly 12C5 hands, since we need 5 pieces of paper from 12 things.  12C5 = 12!/(7! * 5!) = 12*11*10*9*8/(5*4*3*2*1) = 792.

36/792 = 1/22 = the probability that we want.

Now, how do we get to there from where you were?  Your problem was that you made us take our triple before our double.  In fact, we can take the 5 pieces in any order.  This means we need to multiply your answer, 1/220, by the number of ways of mixing a double in with a triple.  How many of those are there?  Well, we need to pick 2 places for the double from 5 -- thus there are 5C2 = 10 possibilities.  So we need only multiply your answer by 10, and (not coincidentally) we will end up with exactly the same answer of 1/22.

Read my message to Leeyuhang on what the probability he calculated was.  The 5C2 is the number of ways to "mix" the three-of-a-kind in with the pair.


Ok, so I got this assignment in class and I need help (dont worry I already worked it out from help of a genius friend, feel free to check it) but his help (his work) is all jumbled up on my paper so I don't know what he did in what order. So could you guys help?

Blah Blah Blah (Its irrelevant stuff)

When Niles proudly announced his plans to Frasier, he said "My new pool will be 4 meters wide."
That's nothing!" exclaimed Frasier. "I'm building one, too, and my new pool will be 6 meters longer than yours. And the width of my pool will be the same as the length of yours.

1. Find the area of Frasier's new pool. Express your answer in simplest radical form- no rounded decimals.
I got 44+ 20 root 7

2. The average depth of Frasier's new pool will be two meters. Find the average depth Niles would need in his pool for the two pools to hold the same amount of water. Again give your answer in radical form. 4+root 7

I am not sure if my answers are right because I have no idea how to check my answers from his work.

Thank you guys

what is the shape of each pool? a circle? a rectangle? there is no way to find anything if the shape is not given. I'm assuming it's a triangle?

Wouldn't it be volume? But in this case probably not because it only gives you a part of the lengths.

I don't think you put enough information for us to solve this problem...

Probably the "Blah Blah Blah" is actually stuff we have to know in order to solve the problem :maybe:

Yes, but I think that wasn't part of the problem... considering the fact he numbered them.

maybe it says the shape of the pool, because that's the most important part of the problem (aside from the numbers)

Isn't that what you just said in the previous post >.< Just saying...

Can't one assume that they're both rectangles, because we're given length and width?
Otherwise, I don't see how one can solve it.

Couldn't it possibly be some other shape like a hexagon? You can find it's length and width... hah...

Oh oops sorry, I forgot to say that by common sense the pool is usually a rectangle so that is what I assumed. The pool is a rectangle sorry about the confusion! FYI the blah blah blah wasnt saying about the rectangle

How about giving us the "blah blah blah" instead of leaving us wondering if it has any useful info necessary to solving this problem?

in addition, since it's a rectangle, what is the other dimension (aside from the $ 4$ feet width)?

Rofl, read what I wrote. It gives you the other dimensions.

Blah BLah BLah= Niles and Frasiers are brothers who have competed with each other all their lives, and continue to do so as adults. When Niles decided to put a rectangular pool in his backyard this summer, Frasier learned of his plans ad decided he had to puta pol in his yard, too. To annoy Niles as much as possible, Frasier plans to build the same shape rectangle as his brother (the ratio of length to width is identical),but of course he will make it bigger.

When Niles proudly announced his plans to Frasier, he said "My new pool will be 4 meters wide."
That's nothing!" exclaimed Frasier. "I'm building one, too, and my new pool will be 6 meters longer than yours. And the width of my pool will be the same as the length of yours.

1. Find the area of Frasier's new pool. Express your answer in simplest radical form- no rounded decimals.
I got 44+ 20 root 7

2. The average depth of Frasier's new pool will be two meters. Find the average depth Niles would need in his pool for the two pools to hold the same amount of water. Again give your answer in radical form. 4+root 7


Adrian, Harun i Jelena su se udružili, skupili novac i kupili tri knjige - jednu od Meše Selimovića, jednu od Zorana Ferića i jednu od Miljenka Jergovića. (Melkior nije sudjelovao u kupovini jer je kupio samo za sebe još jedan primjerak Balkanskog špijuna.) Svaka knjiga ima točno 200 stranica. Budući da su u stisci s vremenom jer se bliži njihov nastup na najvećem balkanskom kvizu ljubitelja lijepe književnosti, na kojem je, ironije li, glavna nagrada zlatom uvezeno izdanje Problem Solving Strategiesa od Engela, odlučili su pomoći jedno drugome.

Svatko od njih može sudjelovati u čitanju točno dvije od tri knjige, budući da žive u različitim gradovima, a poštanske usluge ovih dana nisu jeftine. Svatko može pročitati neke stranice (možda niti jednu, možda sve) iz svake od te dvije knjige. Poznato je da Jelena kategorički odbija čitati išta od Selimovića jer je sve to već pročitala te joj je već dosta i derviša i smrti. Također, Adrian je u zadnje vrijeme ponešto lud, pa čita samo parne stranice knjiga.

Na koliko načina Melkior može savjetovati Adrianu, Harunu i Jeleni koje stranice kojih knjiga da svatko od njih pročita tako da, kad si međusobno prepričaju što su pročitali, svi znaju potpunu radnju sve tri knjige?
-----------------------------------

Dobro, zadatak je zaista lagan i siguran sam da ne bi predstavljao nikakav problem u rješavanju niti jednom od troje naših protagonista (četvrti ga je smislio, pa ni njemu vjerojatno ne predstavlja nikakav problem jer ima pred sobom rješenje), jer je čisto računski.

Planiram ga dati sutra za zadaću jednom od mojih četvorice protégéa, pa sam mislio, možda će nekoga ovdje zainteresirati.

Ako netko ima neki zadatak koji je smislio, a da ga ne čuva za nikakvu svrhu kao natjecanje ili nešto, zanimalo bi me da ga tu stavi. Naime, svi zadaci koje ja smislim u različite svrhe su, bar se meni čine, dosta slični, pa me zanima kako razmišljaju drugi ljudi koji povremeno smisle neki zadatak.

Većinu zadataka koje smislim čuvam za neko natjecanje (pa makar to bio i MNUB :D), a ostali nisu ni približno toliko zanimljivi kao ovaj tvoj zadatak :D

Evo, recimo

http://www.mathlinks.ro/viewtopic.php?t=216751

http://www.mathlinks.ro/viewtopic.php?t=216764

Imam jedan zadatak kojeg sam smislio, ali mu ne znam rješenje. Dražić mi se javio prije par minuta i rekao da ga je riješio... poslao mi je neki početak rješenja i to je imalo smisla, pa pretpostavljam da ću mu vjerovati  :D. Uglavnom, evo zadatak:

http://www.mathlinks.ro/viewtopic.php?p=1337683 (btw, moj 200. post!)

Neka je $ (a_n)$ niz prirodnih brojeva zadan na ovaj način: $ a_1 \equal{} t$, a $ a_n \equal{} i(a_1) \plus{} i(a_2) \plus{} ... \plus{} i(a_{n \minus{} 1})$ za $ n > 1$. $ i(x)$ označava broj jedinica prirodnog broja $ x$ u dekadskom zapisu. Hoće li, bez obzira na izbor $ t$, nakon konačno mnogo članova ovaj niz postati konstantan?

Inače, djuro, tvoji zadaci (barem ono što sam imao priliku vidjeti) jako mi se sviđaju - ja se ne mogu suzdržati od toga da pokušavam poopćiti zadatak, pa dobijem nešto s hrpom kvantifikatora i ne previše smisla...  :D

[quote="mornik"]
Inače, djuro, tvoji zadaci (barem ono što sam imao priliku vidjeti) jako mi se sviđaju - ja se ne mogu suzdržati od toga da pokušavam poopćiti zadatak, pa dobijem nešto s hrpom kvantifikatora i ne previše smisla...  :D[/quote]

Hvala! Nije mi previše jasno to sa poopćenjima - misliš li na poopćenja mojih zadataka ili općenito poopćenje kao metoda izrade zadatka?

Ne, ne, ne mislim na poopćavanje tvojih zadataka - nego, kad smislim zadatak, čini mi se kako je prelagan i onda ga odlučim nekako poopćiti i time učiniti težim.

E, vidiš, onda dobijem nešto što uglavnom izgleda ovako "Neka je $ n$ cijeli broj [i](nap. u početku je bio prirodan)[/i]. Dokaži da za bilo koji prirodan broj $ k$ [i](nap. u početku je bilo $ k\equal{}2$)[/i] postoji nenegativan paran broj $ q$ [i](nap. u početku bilo $ q\equal{}8$)[/i]..." Shvaćaš?  :lol:

Još jedan iz moje radionice, koji po slobodnoj procjeni dugo neće biti riješen :D

http://www.mathlinks.ro/viewtopic.php?t=245167


trigonometry integration calculus algebra polynomial function Euler

here is a nice one i think:
define f(a/b)=a*b over $Q^{+}$ where a/b is irreducible.
Compute $\sum_{q \in Q^{+}} f(q)$ ...

Maybe I'm missing something here, but does that converge? Don't we have $f(n)=n,\ \forall n\in\mathbb N^*$?

Actually, a similar problem (which is what I think you meant) has been [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=5728]posted before[/url]. That topic is locked, but I kind of remember it being posted again and being solved.

of course it is the same pb and I've forgotten the square  :D . It is
sum(f(q)^2,q>0 rational)

Do we need to calculate $\sum_{(m,n)=1}\frac{1}{(mn)^2}$?

In that old thread it says that it should be $\frac 52$, but I'm getting something else, and I wonder what I'm doing wrong.

It can be written as $\sum_{n\ge 1}\frac{2^{t(n)}}{n^2}$, where $t(n)$ is the number of prime factors of $n$. This, in turn, can be written as (this is where I think there should be something wrong) $\prod (1+\frac 2{p_i^2}+\frac 2{p_i^4}+\ldots)$, where $p_i$ are the primes. Rewrite that as $\prod(\frac{2}{1-\frac 1{p_i^2}}-1)=\prod(\frac{1+\frac 1{p_i^2}}{1-\frac 1{p_i^2}})$. This is $\frac {16}6=\frac 83$, I think.

:? Where did I go astray?

I have came to $\frac{\zeta^2(2)}{\zeta(4)}$. But I don't remember $\zeta(4)$ :( Do you know it?

It's $\frac{\pi^4}{96}$. Your result is the same as mine.

Does anybody remember how to calculate $\zeta(4)$? I think there are at least two ways: using Fourier series and complex analysis.

Do you assume we know $\zeta(2)$? It's easy to calculate if we know $\zeta(2)$.

As I remember, for $\zeta(2)$ we need to consider Fourier series for $\frac{\pi-x}{2}$ or something like that.

${\zeta}(2)=\frac{{\pi}^2}{6}$,if anyone is interested I can post the method of [b]Euler.[/b]

[quote="LevonNurbekian"]${\zeta}(2)=\frac{{\pi}^2}{6}$[/quote]
Of course, it is very well known fact.

[quote]if anyone is interested I can post the method of [b]Euler.[/b][/quote]
I think you mean
\[\sin x=x(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{4\pi^2})(1-\frac{x^2}{9\pi^2})...\]
Then you need to justify this identity.  ;)

[quote="grobber"]It's $\frac{\pi^4}{96}$.[/quote]
Actually, I believe $\zeta(4)$ is $\frac{\pi^4}{90}$.

[quote="grobber"]Oh.. That's why I wasn't getting $\frac 52$! :) Thanks! Indeed, it is $\frac{\pi^4}{90}$.[/quote]
Hmm.. Bad memory. You need vitamins, grobber!  :lol:  :lol:  :lol:

It's not bad memory: I never remember stuff like this because I don't want to. I read it in a book, and I understood $96$ instead of $90$. I guess you could say the reason is bad eyesight :D.

All this talk of $\zeta(4)$: here's a Fourier series calculation using Parseval's identity.

If $f\in L^2(\mathbb{R}/\mathbb{Z})$, then we can write $f$ as $f(x)=\sum_{k=-\infty}^{\infty}\hat{f}(k)e^{2\pi ikx}.$ Parseval's identity is the basic fact about an expansion by an orthonormal basis:

$\int_0^1|f(x)|^2dx=\sum_{k=-\infty}^{\infty}|\hat{f}(k)|^2.$

Let $f(x) = \frac12-x$ on $(0,1).$ We can calculate $\hat{f}(k)=\frac1{2\pi ik}$ for $k\ne0$ and $\hat{f}(0)=0.$

Now let $g$ be the antiderivative of $f$. That means that $g(x)=C+\frac{x}2-\frac{x^2}2.$ We choose $C$ such that the integral of $g$ is zero, from which we obtain that $C=-\frac1{12}.$

For $k\ne0,\hat{g}(k)=\frac{\hat{f}(k)}{2\pi ik}=\frac1{(2\pi ik)^2}=\frac{-1}{4\pi^2k^2}.$

Since we set the integral of $g$ equal to zero, $\hat{g}(0)=0.$

Now apply Parseval's identity to $g$:

$\sum_{k=-\infty}^{\infty}|\hat{g}(k)|^2= \sum_{k\ne0}\frac1{16\pi^4k^4}=\frac2{16\pi^4}\zeta(4)$

$\int_0^1|g(x)|^2dx= \int_0^1\left(-\frac1{12}+\frac{x}2-\frac{x^2}2\right)^2dx= \frac1{720}$

Thus, $\frac1{8\pi^2}\zeta(4)=\frac1{720}$ or $\zeta(4)=\frac{\pi^4}{90}.$

If we keep integrating this polynomial, choosing the constant at each step so that the function at that step has integral zero, we can extend this method to calculate zeta at any positive even integer. So can several other methods, including Euler's infinite product method which was alluded to earlier in this thread. I believe Euler himself calculuted $\zeta(26),$ which is quite a considerable feat of calculation. The first few values are:

$\zeta(2)=\frac{\pi^2}6,\zeta(4)=\frac{\pi^4}{90}, \zeta(6)=\frac{\pi^6}{945},\zeta(8)=\frac{\pi^8}{9450}.$

This approach is known for me. Can anyone obtain the same result using complex analysis (i.e. residue theory)?

[quote="Myth"]Can anyone obtain the same result using complex analysis (i.e. residue theory)?[/quote]

Let $f(z)=\frac{\cot z}{z^4}.$ The function $\cot z$ has simple poles of residue $1$ at every integer multiple of $\pi.$ Therefore, $f(z)$ has simple poles of residue $\frac1{\pi^4k^4}$ at $z=k\pi,k\ne0.$

At the origin, $f(z)$ has a pole of order $5.$. We employ long division of the power series for cosine and sine to find the Laurent series for $f(z)$:

$f(z)=z^{-5}-\frac12z^{-3}-\frac2{90}z^{-1}+\cdots.$

Thus, the residue of the function at $0$ is $-\frac2{90}.$

Integrate around a square whose vertical sides are given by $|x|=\left(N+\frac12\right)\pi.$ This splits between the poles. We can show that $\cot z$ is bounded on that contour, so that $f(z)=O(N^{-4})$ on the contour. This in turn means that the integral is $O(N^{-3})$ and goes to zero as $N\to\infty.$ Since the integral can be computed from the sum of the residues, we have the following:

$\lim_{N\to\infty}-\frac2{90}+\sum_{|k|\le N,k\ne0}\frac1{\pi^4k^4} =0.$

This can be simply rearranged to $\zeta(4)=\frac{\pi^4}{90}.$

To compute $\zeta(6), \zeta(8),\dots$, we merely need to continue the long division for calculating the Laurent series of $\cot z.$ However, the method provides no information if we divide by an odd power of $z.$


$AO1,AO2$ meets circles $(O1)$ and $(O2)$ at $B,C$
$(O1)$ and $(O2)$ meets at $A,F$
We have $\angle AFC=\angle AFB=90$ so $B,F,C$ are collinear
$AM$ meets $BC$ at $K$. We see $MA=MK$ $(1)$ and $KB=KC$
Otherwise $BE//CD$ (both $\bot ED$) and $NE=ND$ so $KN//EB//CD$
So that $KN\bot ED$. And with (1) we have $NM=1/2AK=AM$ $q.e.d$

Circle $(O1)$ and $(O2)$ meets again at $F$
We have $\angle FDA=\angle FO2O1, \angle FED= \angle FO1O2$ so triangles $FO1O2$ and $FED$ are similar $(1)$
And because $M$ is the midpoint of $O1O2, N$ is the midponit of $ED$ $(2)$ we have $\angle NFD=\angle MFO2$
And then $\angle NFM=\angle DFO2$. Also because of (1) and (2) $FM/FN=FO2/FD$
This means triangles $FMN$ and $FO2D$ are similar
So $FM/MN=FO2/DO2=1$.Thus $MA=MF=MN$ $q.e.d$

[quote="simonlin_bearer"]Let A be an intersection point of two circles C_1 and C_2 with centers O_1 and O_2 respectively. A line through A intersects C_1 at E and C_2 at D. Let M and N be the midpoints of the segments O_1O_2 and ED respectively. Prove that MN=MA[/quote]

Also check out my post at http://www.mathlinks.ro/Forum/viewtopic.php?t=72363 . Please find suitable titles for your posts! For instance, "two intersecting circles and two midpoints" would already be quite good.

  Darij

[quote="simonlin_bearer"]Let $A$ be an intersection point of two circles $C_{1}$ and $C_{2}$ with centers $O_{1}$ and $O_{2}$respectively. A line through $A$ intersects $C_{1}$ at $E$ and $C_{2}$ at $D$. Let $M$ and $N$ be the midpoints of the segments $O_{1}O_{2}$ and $ED$.Prove that $MN=MA$[/quote]

  If $B,C$ are the projections of $O_{1}, O_{2}$ on line $ED$ , then $NA+2AB=NC+CD$ or $AC-CN+2AB= NC+CD$ or
   $2AB=2NC$ , that is $AB=NC$. If $P$ is the projection of $M$ on $ED$ , then $P$ will be the midpoint of $BC$ and since   $AB=NC$, we have also that $PA=PN$. This gives  that triangle $MAN$ will be isosceles at $M$.

   [u]Babis[/u]

[u] Note[/u]

[i] This solution is based on an idea used by Vittasko in a solution of a similar problem.[/i]


So guys, what do you think you got on Pascal/caley/fermat?

I personally had no time left, but to guess and I eliminated 3 options for sure, and when so I just picked either of the 2 remaining choices... And after the contest, I find the answer I've chosen is wrong -.-

Didn't get the last three questions, which was horrible, so about 130ish

Can someone post the alphabetic solutions for 1 - 25 for each test so we can score them.

i compared with friends, i think i got 150 on pascal.
the only one that is sort of on the edge is 25. (is number 25 d?)

Heres what I got on the Fermat
1
[hide] D
[/hide]
2
[hide] E
[/hide]
3
[hide] C
[/hide]
4
[hide] B
[/hide]
5
[hide] A
[/hide]
6
[hide]A
[/hide]
7
[hide]D
[/hide]
8
[hide]D[/hide]
9[hide]C[/hide]
10[hide]E[/hide]
11[hide]D[/hide]
12[hide]C[/hide]
13[hide]A[/hide]
14[hide]E[/hide]
15[hide]B[/hide]
16[hide]C[/hide]
17[hide]B[/hide]
18[hide]B[/hide]
19[hide]A[/hide]
20[hide]A[/hide]
21[hide]D[/hide]
22[hide]E[/hide]
23[hide]D[/hide]
24[hide]D[/hide]
25[hide]
I didnt get it but Newanda posted a solution and I think its correct. If so then its C
[/hide]

How can you get D on #8? I just got C though :(

I wasn't sure whether it was C or D because  the question never said different songs. I wasn't sure if "maximum number of songs from his collection" implied that they were different songs or they could be the same.

What do you meant by different songs?? I interpret this problem as using those songs of 2 kinds: 1 last for 3 minutes and 1 last for 5 minutes. So, the maximum songs will occur when we use as many 3 minutes songs as possible+some songs last for 5 minutes. In this case, it must be 56. How can you get such 60?Because if you play 60 songs, then there are at least 10 songs lasting for 5 minutes. So, the maximum minutes you now have is $ 180\minus{}50\equal{}130$. So the maximum number of songs lasting for 3 minutes are 130/3=43.333, which is approximate 43. So the number of songs now are just 10+43=53, which is less than 56.(remember that each song values the same, so the more songs lasting 5 minutes you hear, the less songs lasting 3 minutes you hear).

What I mean is can't you play 60 three minute songs (playing 10 twice)? I'm not sure if you are allowed to play a song twice or not.

There was a question like that on caley, about the 3 and 5 min songs and I'm pretty sure you can't repeat the songs, well I think I tried that possibility but it wasn't in the answer so, I think.. >.> but I got 6 x 5min songs

can anyone post the answer for the cayley competition, my brother would like to know how he does

Why won't the cemc post the answers and questions yet, if they have unofficial results???

Here's mine, no guarantees though.

1.A
2.E
3.C
4.D
5.A
6.D
7.E
8.B
9.D
10.A
11.C
12.C
13.E
14.D
15.A
16.B
17.B
18.B
19.D
20.B
21.E
22.E
23.A
24.C
25.C

and i guessed those last two.

i am pretty sure the answer for no.24 is 5/12, but i couldn't manage to do 25 as well


[b]Mathematical Physics Equations: Exact Solutions & Exact Methods - Useful Links[/b]

*Mathematical Physics Equations - Index
(heat equation, wave equation, Boussinesq equation, Klein-Gordon equation, Korteweg-de Vries equation, Monge-Ampere equation, Navier-Stokes equations, Schrodinger equation, sine-Gordon equation, etc.)
http://eqworld.ipmnet.ru/en/solutions/eqindex/eqindex-mphe.htm

* Linear Equatins of Mathematical Physics - Exact Solutions
heat equation, wave equation, Laplace equation, Poisson equation, Helmholtz equation, etc.)
http://eqworld.ipmnet.ru/en/solutions/lpde.htm

* Nonlinear Equatins of Mathematical Physics - Exact Solutions
(Boussinesq equation, Klein-Gordon equation, Korteweg-de Vries equation, Monge-Ampere equation,
Navier-Stokes equations, Schrodinger equation, sine-Gordon equation, boundary layer equations,
Fisher equation, FitzHugh-Nagumo equation, etc)
http://eqworld.ipmnet.ru/en/solutions/npde.htm

* Nonlinear Equatins of Mathematical Physics - Exact Methods
 (Nonclassical and conditional symmetries, Methods of Generalized
and Functional Separation of Variables, Differential Constraints Method, Singularity Analysis)
http://eqworld.ipmnet.ru/en/methods/meth-pde.htm

* Nonlinear Systems of PDEs - Exact Solutions
(Systems of Reaction-Diffusion Equations, Systems of PDEs of General Form)
http://eqworld.ipmnet.ru/en/solutions/syspde.htm

Thanks a lot for the links....a lot of them seem pretty good.....and do you know any links where we get to find an explanation and some problems based on it as well ??


Does interpretations of the derivative in AP cal test?

What do you mean by that? Do they ask about different ways to calculate derivatives? Go to [url]http://collegeboard.com[/url] to find out...