I'm new to this site so I apologize if this is under the wrong topic.

I was hoping people who knows about PROMYS can give me some insight into the pogram.

I'm in the middle of filling out the application. I did some research and found out that it only deals with number theory. Would I be at a really big disadvantage if I have absolutely no background on the subject if I somehow manage to be accepted into the pogram?

I have also only manage to answer a few of the questions and am struggling with a good number of them. From the posts I've been reading here about those who answered the majority of the questions already, I kind of feel that I might not on the same level as most of the people going to PROMYS. I'm pretty sure I can figure these questions out or at least partially answer them, but I can see that I'm little slower at solving these types of problems.

I guess what I'm trying to ask is should I bother applying? Despite my lack of knowledge on number theory and this slight struggle on the application, I am really interested in the subject and am more than willing to do the work given to me even if it might take me longer than others. I was hoping if anyone who had experience with PROMYS can tell me more about life in the pogram and whether or not you think the six week was worth the money.

[quote="DB-Kage"]Would I be at a really big disadvantage if I have absolutely no background on the subject if I somehow manage to be accepted into the pogram? [/quote]
Nope.  PROMYS starts from the very beginning; you'll learn all the number theory you need to know there if you can keep up with the problem sets.

[quote="DB-Kage"]I have also only manage to answer a few of the questions and am struggling with a good number of them. From the posts I've been reading here about those who answered the majority of the questions already, I kind of feel that I might not on the same level as most of the people going to PROMYS. I'm pretty sure I can figure these questions out or at least partially answer them, but I can see that I'm little slower at solving these types of problems.[/quote]
This might be a background issue and not a problem with your problem-solving ability.  Keep trying and apply anyway!  You'll never know otherwise.

[quote="DB-Kage"]I was hoping if anyone who had experience with PROMYS can tell me more about life in the pogram and whether or not you think the six week was worth the money.[/quote]
I'll be brief:  [b]PROMYS made me decide I wanted to be a mathematician.[/b]  And my only regret is not doing more while I was there.

yeah im done but the app is only due in may. t0ra ive done 9, do you still think it is worthwile to wait and solve that last problem?

Thanks t0rajir0u
I think I'll at least try it and see what happens from there.

Oh yeah, do you know the average age range and grade level of the kids who go there for their first year?


Find all real numbers $x$ such that $x^n + x^{-n}$ is an integer for all $n \in Z$.

Let $y_n=x^n+x^{-n}.$ We have $y_{n+1}=y_ny_1-y_{n-1}$. Therefore if $y_1$ is integer ($y_0=2$),then $y_n$ are integer for all n. $x=\frac{n\pm \sqrt{n^2-1}}{2},|n|>1$.

Shouldn't this be
\begin{eqnarray*} x & = & \frac{n \pm \sqrt{n^2 - 4}}{2} \end{eqnarray*}
Otherwise it's very nice !


Remove the upper-left and lower-right corner squares from an 8 x 8 chessboard. Show that the resulting board cannot be covered by 31 dominoes. (A domino consists of two squares with an edge in common.)

[hide]Since the two removed squares were on the same diagonal, they are the same color.  A domino covers two adjacent squares, which must have different colors.  Therefore after placing 30 dominoes, there will be two remaining squares that have the same color, so these two cannot be covered with one domino and it is impossible to cover the board with 31 dominoes.[/hide]


How do you evaluate sums like
\[\mathcal{S}(a,b)=\sum_{k=1}^{\infty}\frac{1}{k^{2}+a\sqrt{k}+b}\]
with $(a,b)\neq (0,0)$. Probably this is hard in general but please post some special cases
and methods.

Easy special case:  $a = 0, b =-r^{2}$.  We have an easy telescoping sum.

Thanks a lot. Methods?

sos440: Is your first sums correct? what if $x$ approaches zero?

[color=red]Please post more[/color] cases also with nonzero $a$.

There is a removable singularity at $x=0$ in sos440's expression; the two parts grow at essentially the same speed.

What makes you think there are any nice answers with $a\neq 0$?

:oops:  Sorry but I dont, thats why I ask. I was just wondering if for example the
series
\[\sum_{i=1}^\infty\frac{1}{i^{2}+\sqrt{i}}\]
could be evaluated, since I think I have seen something like that, or perhaps it was a
convergence question.


Now that the season is over, how far did you all go in MATHCOUNTS this year?

[New option added by solafidefarms]

Made states and countdown :) , but didn't get to nationals.

didn't make any countdowns, but I got to state.

I think "didn't make anywhere" at nats is unfair. You are ranking 13th the same as 228th and 13th at Nationals is a very respectable place. You should do like, made top 118th (not 114th :P) at Nationals and another saying top 57 or something.

I agree with Ignite168, especially since you didn't make it to nats ZMaster, no offense or anything.

I made it to nats, just missed countdown at nats, got 14th place.

I'm sorry it sounded offensive.  I was just going to bed and was really tired and was rushed by my parents to go to bed.  Don't take it as an offensive comment.  Be glad you made nats.

Heheh... I got ... 72nd at Nationals!!! Yay me, I guess.

[quote="k8reindeer"]what exactly is the signifgance of 118th place?[/quote]
heh. heh. heh.
NO SIGNIFICANCE AT ALL.
 :D

Heh. that's the rank 24 would have gotten you :P.

This year was full of surprises (No, not connecticut in particular).

1. Kevin Yang and Phillip Bauer

2. Me getting a 28 (lol, not really).

3. MI, Cali, and NY all 10-15

4. Daesun not being on the polls and winning (partially because of the ban)

5. :?:

Neil Gurram:I made it to nats, just missed countdown at nats, got 14th place.

Neil i told you that you came in around 13th....actually i said 13th exactly but 14th is pretty close.
reword: very close.
112th aint bad.[/quote]

[quote="Klebian"]WAIT WHAT?
WHO ADDED THE NEW OPTION?
That's just not cool. Can you change it to above say... 120th?
 :P[/quote]

Now do I satisfy you? :P

XD Owned.

I failed my goal at Nationals, I got a 30 instead of a 47 :(

[quote="Waxing Rhapsodic"]I got 132nd at nats with a score of 892.

I guess their socring is a bit inaccurate.[/quote]

What would a 23 have gotten you then?

[quote="Zmastr"]I didn't ask anyone to change the poll options,  I guess you guys just took the liberty of changing it without telling me. :P.  But thanks anyways solafidefarms, because I didn't know exactly what to do to change the options anyways. :)[/quote]

Only moderators can do it, but now a lot of people that did 118th or better voted below 118th because of it, oh well.

[quote="lotrgreengrapes7926"][quote="Ignite168"][quote="lotrgreengrapes7926"]50th place. Yay. Not quite :first: but oh well. :)[/quote]

We were sitting in front of you guys for the awards banquet. Are you Ryan Gao? If so, do you talk?  :huh: Afterwards, we were all amazed about how you remained so quiet and barely moved during the whole thing.[/quote]

Umm...Thank you? Yes, I do talk, just not around smart people.  :P What state are you from? We're we sitting at the same table or something?[/quote]

Connecticut, we were in front of you guys :P I was the one in the maroon polo shirt sitting next to the coach and the one with a polo shirt down to his knees. (That's another story)

[quote="Zmastr"]Were tickets available to go to nats?  I hope not, because then I would've gone.  :([/quote]

You could've registered so yes, you could've gone.


Let z be a complex number.If       |z+(1/z)|=3 then find the maximum and minumum value of z.

That should probably be, "the maximum and minimum value of |z|."

The value of $|z|$ is minimum or maximum, if the triangle $0, z, \frac 1 z = \frac{z*}{|z|^2}$ degenerates to a line, i.e., if $z$ is real or purely imaginary.

If $z = x$ is real,

$\left| x + \frac 1 x \right| = 3$

and we can assume $x > 1$. Then

$x + \frac 1 x = 3$, $x^2 - 3x + 1 = 0$, $x = \frac{3 \pm \sqrt 5}{2}$,

take the minus sign for the minimum $|z| = x$.

If $z = iy$ is imaginary,

$\left| iy + \frac{1}{iy} \right| = \left| iy - \frac i y \right| = \left| y - \frac 1 y \right| = 3 $

and again, we can assume $y > 1$. Then

$y^2 - 3y - 1 = 0$, $y = \frac{3 \pm \sqrt{13}}{2}$,

take the plus sign for the maximum $|z| = y$.

In fact, one can show that the solution obtained where $z = yi$ leads to the maximal $z$. To do this,
[hide]you could use polar form $z = re^{i\theta}$ and then make some geometric arguments.[/hide]


When you're writing proofs or solutions do you have to write them completely forward, will you get points taken off?

No. I've seen solutions that have gotten 7s which use stuff like "Claim: x + y + z = x^x."

Though I think that unless you use claims, backward solutions typically don't make the grader happy, which makes minor mistakes less minor.


for example if you looked for x as 1
$n^2=1+n$
$n^2-n-1=0$
n solves for something like $.5+\sqrt(1.25)$

chess64's problem

using the same method of squaring and plugging in the original equation,

I got

[hide="answer"]6[/hide]

it's probably wrong....

[quote="superman"][quote="chess64"][quote="pgpatel"]how would you solve sqrt(x+sqrt(x+sqrt(x+...[/quote]

If $N=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$, then square both sides.

$N^2=x+\sqrt{x+\sqrt{x+\cdots}}$. Substitute.

$N^2=x+N$. Then solve the quadratic equation for the positive root.

Now try this. Solve for $x$: $3=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$.[/quote]

when u give huge hints like that why dont u just answer it?[/quote]

I was just telling him how to do it.

[quote="chess64"][/quote][quote="superman"][quote="chess64"][quote="pgpatel"]how would you solve sqrt(x+sqrt(x+sqrt(x+...[/quote]

If $N=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$, then square both sides.

$N^2=x+\sqrt{x+\sqrt{x+\cdots}}$. Substitute.

$N^2=x+N$. Then solve the quadratic equation for the positive root.

Now try this. Solve for $x$: $3=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}$.[/quote]

when u give huge hints like that why dont u just answer it?[/quote][quote="chess64"]

I was just telling him how to do it.[/quote]

He's only trying to help

[quote="nat mc"][quote="Xantos C. Guin"]:D Good point.
$25=\sqrt{p+\sqrt{p+\sqrt{p+\sqrt{p...}}}}$
Solve for p[/quote]
600[/quote]
Good Job!

What if 1 = $\sqrt{x+\sqrt{x+...}}$
what's $x$?

[quote="nat mc"]no solution.
$1=\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x...}}}}$
$1=x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x...}}}}$
$1=x+1$
$x=0$

But that onviously can't work[/quote]

That's because $N^2=N \Rightarrow N^2-N=0$.

besides negatives, I think 1 is the only number that doesn't work

I used about 2 hours on a plane ride to china (resulting a miss of the state competition) to figure out this problem. I came close to 1, but that was with a calculator and severe endurance. I don't even know what the number was anymore.

now what if we were to do:

$x = \sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\sqrt{2\ldots}}}}}}}}$

same concept but you end up with a quartic. The solutions to the quartic were quite odd but some are easy to guess. as I found them. Hint: The actual value of x is the positive irrational solution to the quartic.

what if we were to do and very small value for $1=\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x...}}}}$ and plug it in. I'll use 0.01

0.01 makes it 1.006633...

0.0001 makes it1.000066663333...

0.00000000001 makes it very close to 1. However it will not be 1. Thus, we infer that as x approaches 0, $\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x...}}}}$ approaches 1 without quite getting there.

by the way, to do it on a graphing calculator is easy. For example, on a TI-84 i typed:

$0.00000000001->x$
$\sqrt{\text{ }}(0.00000000001+\sqrt{\text{ }}(x))->x$

and repeatedly pressed enter

I like what I see, hamster1800. You explanation is very well said.

What I found, that for $\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+...}}}}$
The answer is one if you do the old squaring thing.
$N^2 - N - 0 = 0$
N = 0
N = 1.

In this case, zero is the answer that seems to make sense.

[quote="Hamster1800"]

0.00000000001 makes it very close to 1. However it will not be 1. Thus, we infer that as x approaches 0, $\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x...}}}}$ approaches 1 without quite getting there.
[/quote]

according to what is written there, it looks like that when x is 0, the sum [i]is[/i] one.

no. It's the same concept as that while x approaches 0, $x^x$ approaches 1, but doesn't get there with $0^0$ because it's undefined.

The only difference is that in this case, is that instead of becoming undefined, it becomes 0. Here's why:

$x = \sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\ldots}}}}$

$x = \sqrt{0 + x}$

$x^2 = x$

$x = 0, 1$, however, there is no value to create a 1, so therefore it is 0. Try it with a  graphing calculator:

$0 ->x$
$\sqrt{\text{ }}(0 + \sqrt{\text{ }}(x))->x$

and repeatedly press enter.

you could also do it this way:

$x = \sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\ldots}}}}}$

square it and you get

$x^2 = \sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\sqrt{0+\ldots}}}}}$

continue squaring it and you'll get:

$x^\infty = 0$ so x can obviously not be 1. (i know i shouldn't use $\infty$ but it's the easiest way to express it)

This is a pretty useful mathcounts/AMC10 trick. There were several problems like this. Is that the fastest way, or is there an even faster formula(sorry AoPS instructors) that would work well?

If we have the problem:

$N = \sqrt{x+\sqrt{x+\sqrt{x + \cdots }}}$

And we know $N$: [hide]
$x=N(N-1)$[/hide]

And we know $x$: [hide]
$N^2=x+\sqrt{x+\sqrt{x+\sqrt{x + \cdots }}}$
Substitute: $N^2=x+N \iff 0=N^2-N-x$. Quadratic formula:
$N= \frac{1 \pm \sqrt{1-4(1)(-x)}}{2}=\frac{1 \pm \sqrt{1+4x}}{2}$

So $N=\frac{1 \pm \sqrt{1+4x}}{2}$. See which solution is positive. Not a formula worth memorizing, though  ;)
[/hide]


16)

Vinuta and Katy are standing on the points $ (3,6)$ and $ (3,\minus{}10)$ respectively. They decide that they want to meet at some point on the ellipse $ \frac{(y\plus{}2)^2}{289} \plus{} \frac{(x\minus{}3)^2}{225}\equal{}1$, so they both walk in a straight line toward that point. What is the sum of the distances that Vinuta and Katy must walk before they are reunited?

What I know: an ellipse is like an oval shape, with its center at the origin. That's about it. Oh, and Johannes Kepler furthered Tyco Brahe's data to find that the planets rotate around the sun in the shape of an ellipse.

22) Find the length of the latus rectum of the conic section with equation $ \frac{(x\plus{}1)^2}{\sqrt{2}} \plus{}2(y\plus{}1)^2\equal{} \sqrt{2}$

I have no idea how to do this. I wonder if there are very good websites to learn algebra 2 level conic sections.

Help would be awesome!

[url=http://www.komal.hu/cikkek/dandelin/dandelin.e.shtml]Here is a link.[/url]

[hide="Big Hint for #16"]
Recognize that the midpoint of the two points $ (3,6)$ and $ (3,\minus{}10)$ is $ (3,\minus{}2)$.

Knowing this, substitute the $ y$ variable with $ \minus{}2$. The point on the ellipse with $ (x,\minus{}2)$ (the $ x$ variable is given when we substitute $ y$ in the expression, try finding the value of $ x$ first) is where the two points should meet.

Evaluate the distance of one point to the new point given that's on the ellipse (know that both points are equidistant from the given point on the ellipse).

The distance of two points $ (x_1,y_1)$ and $ (x_2,y_2)$ can be evaluated with this: $ \sqrt{(x_2\minus{}x_1)^2\plus{}(y_2\minus{}y_1)^2}$. [/hide]

16) Just use the definition of an ellipse.
[hide="Answer"]17*2=34[/hide]


calculus
derivative
calculus computations

Fint the equation of the tangent to the curve $ y \equal{} e^x$ that passes through the origin.
[hide="Note:"]If your answer be $ y \equal{} xe^x$, think again! This is a [b]curve[/b] and not a [b]Straight Line[/b] as required. $ ($Refer Graph below$ )$[/hide]

$ f(x)\equal{}e^x$

$ f'(x)\equal{}e^x$

$ L(x)\equal{}f'(y)(x\minus{}y)\plus{}f(y)\equal{}f'(y)x \to f(y)\equal{}yf'(y)$

$ e^y\equal{}ye^y\to y\equal{}1$

so the line $ L(x)\equal{}ex$

hello
[hide="solution"]
any line passing through origin will be in the form y=mx
$ y \equal{} e^x$
$ m \equal{} y' \equal{} e^x$
now $ xe^x \equal{} e^x$
=>$ x \equal{} 1$
thus point on the curve is $ (1,e)$
thus the eq. of tangent is $ y \equal{} ex$
[/hide]
Thank u.

[quote="kabi"]hello
[hide="solution"]
any line passing through origin will be in the form y=mx
$ y \equal{} e^x$
$ m \equal{} y' \equal{} e^x$
now $ xe^x \equal{} e^x$
=>$ x \equal{} 1$
thus point on the curve is $ (1,e)$
thus the eq. of tangent is $ y \equal{} e^x$
[/hide]
Thank u.[/quote]
Kabi, I think you mean $ y \equal{} ex$, don't you.
Thank you.

Hello
sorry by mistake I wrote $ y\equal{}ex$ as $ y\equal{}e^x$.


Prove that every natural numbers can be written as difference of two natural numbers which have the same number of prime factors.

Distinct prime factors, or with multiplicities?

if n is even then n=2n-n
if n is odd ,we consider the factorization of n in prime factors:let p the smallest prime factor greater than 2 that doesnt appear in the factorization of n
p+1 is even so it has 2 as prime factor
all the prime factors of p+1 are factors of n since if r is a factor of p+1 then it is smaller than p so it apears in the factorization of n as we defined p.
So write now n=(p+1)n-pn


$ x \equal{} r \cos \theta$
$ y \equal{} r \sin \theta$
$ r \equal{} \sqrt{x^2 \plus{} y^2}$
$ \theta \equal{} \arctan{y/x}$

That should be all you need.

[quote="sharkman"]The letters x and y represent rectangular coordinates.  Write each equation using polar coordinates (r, theta).

(1) 4x^2y = 1

(2) y = -3[/quote]

[hide="Number 1"]
I'm not sure how to do this one, perhaps it should have been $ 4x \plus{} 2y \equal{} 1$?
[/hide]

[hide="Number 2"]
Since $ y \equal{} r \sin \theta$, we have $ r \sin \theta \equal{} \minus{}3$
[/hide]

[quote="sharkman"]The letters x and y represent rectangular coordinates.  Write each equation using polar coordinates (r, theta).

(1) 4x^2y = 1

(2) y = -3[/quote]

[hide="is this right for #1?"]$ 4r\cos\theta^{2r\sin\theta}\equal{}1\implies 4r\cos\theta\equal{}1\implies r\cos\theta\equal{}\dfrac{1}{4}$[/hide]

Learn LaTeX.  Did you mean $ 4x^2y \equal{} 1$ for #1?

[quote="lingomaniac88"]Did you mean $ 4x^2y \equal{} 1$ for #1?[/quote]

That might be easier to do  :) .

I don't get #H34N1's solution though, can you explain it to me? It seems like the $ 2r \sin \theta$ mysteriously disappeared.

THEANI's solution was: $ a^b \equal{} 1 \implies a\equal{}1$. However, a more correct solution to his interpretation ( (4x)^(2y) = 1 )would read $ a^b \equal{} 1 \implies a\equal{}1; \text{ or } b\equal{}0 \text{ and } a \neq 0$

$ a\equal{}1$ yields the solution he gave, $ b\equal{}0$ yields $ 2r \sin \theta \equal{} 0 \implies \theta \equal{} 0 \text{ or } \theta \equal{} \pi$


Uh..how do you exactly graph $(4\cos{x})^{2}+\cos{x}-3$

The original problme was find the minimum of f(x)=cos(x)+cos(3x)/cos(x)..and I manage to get to that step but Im kind of unclear of the graphing (the minimum is easy to find but how do you graph it?

Let $t=\cos x$ for $t \in [-1,1]$
So,
$4 \cos^{2}x+\cos x-3 = 4t^{2}+t-3 = 4 \left (t+\frac{1}{8}\right )^{2}-\frac{49}{16}$
Therefore, minimum value of f(x) is $-\frac{49}{16}$ for $x$ which satisfies $-\frac{1}{8}=\cos x$

ya, I got that as the minimum fr just -b/2a, then you sub that back in for cos x, but i was just wondering in general, how do you graph this.  When I graph this 4cos^2x+cosx-3, I use x=cos x, so 4x^2+x-3=0, its just a parabola where x is restricted between [-1.1], but however how do you graph this just in terms of the angle measure x, not cos x.


algebra
binomial theorem
Pascal\u0027s Triangle

Find the coefficient of the term from the expansion $ (x^2 \minus{} 2y \plus{} z)^{10}$ that contains $ x^{10}y^3z^2$.

I don't know how to do this one, although I do remember the binomial theorem from algebra 2.

In algebra 2, I remember n! / (n-m)! m!, and Pascal's triangle, but I'm not sure how to use that here, since we only worked with numbers like (x-y)^4, so I'm not sure if the binomial theorem works here? Explanation would be awesome! =)

The relevant theorem is actually a generalization called the [url=http://en.wikipedia.org/wiki/Multinomial_theorem]multinomial theorem[/url].  The counting principle is still the same.

One way of thinking about the multinomial theorem is just treating it as repeated applications of the binomial theorem:  Write $ (x^2 \minus{} 2y \plus{} z)^{10} \equal{} (x^2 \plus{} (z \minus{} 2y))^{10}$ and use the binomial theorem to find the coefficient of $ x^{10}$.  This will leave you with some integer times some power of $ z \minus{} 2y$, and you then use the binomial theorem again to expand the power of $ z \minus{} 2y$ to find the coefficient of $ y^3z^2$.  (In fact, this can be easily used to give an inductive proof of the multinomial theorem.  As t0r notes, the counting proof of the multinomial theorem is basically the same as the counting proof of the binomial theorem.)


I'm sure you guys are as excited as I am.  Roloson and crew played a solid 5 games against the Anaheim Ducks.  ROLOSON FO PRESIDENT!!!

Torres' tip-in goal that clinched the Conference Finals series...that was...beautiful...makes me want to cry...

So guys, who will be the Oil's next prey?  Will it be the physical Carolina Hurricanes?  Or the ub32-rookies Buffalo Sabres?

Calgary's had their chance last year.  Toronto fell short 3 seasons ago.  Now it'll be another 2 weeks of mindless, senseless, breathless, lobsterless, euphoria, as we all watch CANADA'S TEAM MAKE ANOTHER RUN FO DA CUP.  RIGHT GUYS????

WOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO

Psh.  Stupid Flames, they just SUCKED.  Oh well.

Oilers was my next choice for the cup, hehehe.  Probably cuz my parents are HUGE fans, they lived in Edmonton when they won their 5 Stanley Cups!

Go Oilers!!!!

I watched most of the games, and I definitely think that they can win this year. :winner_first:
It's crazy here in Edmonton. An average car would have two Oilers flags. Whyte avenue is always filled with crazy fans. At school, half of the people wear jerseys and all they talk about is hockey, even teachers. :nhl: (very suprised that there is a hockey emoticon :) )

When I was a kid, I lived in Saint John, New Brunswick, and I used to attend Saint John Flames hockey games (an AHL farm team for Calgary), and Dwayne Roloson was their goalie.  I remember him being a pretty solid goalie too.

I really don't keep up with hockey anymore though.  I don't think I've watched a single NHL game this season.  Then again, I really don't watch TV at all nowadays (university thrusts these type of lifestyle changes upon you).

Well I guess that will be one less change I will have to deal with.  I never really watch the TV anymore.  Actually, I don't really like watching the TV.. or movies, either.

No offence to people who love those two things.

I do watch for sports, sometimes.  Specifically, I watch for tennis.

But I am lookin forward to other aspects of my life changing - for the better, of course.  Sorry, I'm a bit off topic.

But ya, as a Canadian, of course I am routing for the Oilers to take it this year.  The Canadian team vs. American team thing is interesting.

[quote="The Master"]The Canadian team vs. American team thing is interesting.[/quote]

That's such an understatement.  If you root for the Carolina Hurricanes/Buffalo Sabres, then you're not Canadian :ninja: .

I used to cheer for Carolina a lot because I felt sorry for them.. :)
They weren't very good last year..

I guess I have to cheer for Edmonton now though..

Maybe they'll have a game while we're at seminar!

Some fans are getting too excited. After the last game, some people caused some trouble.
[url]http://www.cbc.ca/calgary/story/ca-whyte-ave-violence-20060529.html[/url]

I don't watch tv. I like PLAYING sports ( ultimate frisbee comes to mind  :D  ) but watching them? kind of boaring. I think the only times I enjoyed watching a sports match on tv:
    -Some bits of the tour france
    -the olympics
    -when espn covered the mtg world championchips  :lol:  :rotfl:  :D

[quote="Sunny"]Psh copycats!  Whyte Ave is sooooo a copy of the Red Mile in Calgary.  Only...Edmontonians get violent.  Umm, the Calgary people just go around flashing people, lol.[/quote]

Haha you're jealous.

While Calgary lost to the Ducks in 7, the Oilers beat them as if they weren't even a challenge.  WOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO OILERS.

Oh yeah, Buffalo just forced a Game 7.  So I'm hoping whoever wins Thursday, will be super tired on Monday.

[quote="lotus87"][quote="Sunny"]Only...Edmontonians get violent.  Umm, the Calgary people just go around flashing people, lol.[/quote]
 :D Hahahahaha!!
How do Okotokots celebrate!!?  I bet they do something with that rock..  :|[/quote]

 :D Hey, its a pretty big rock!!

Lol, i'm not jealous actually.  Calgary honestly just sucked, lol.  Oilers on the other hand are doing awesome.   :P   I was born in Edmonton.  I can cheer for them!!!   :D

GO CAROLINA GO!!!  :D

I actually only support them to be different from everyone else in my school, but they are a cool team!

They'd better at least beat buffalo.

Carolina won. Hope they are really tired. :D  The first game is on Monday.

:cursing: ARRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRRR!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!   :furious:
Oilers lost. :(  I hope that Roloson :stretcher:  is ok.
[color=white][size=9]I added this so that the I can use the smilies[/size][/color]

al;sdfjl;sdakhg;skahg;shg

Can you believe that???  Ladd took out Roloson!!!  That's totally uncalled for!!!

I guess the only way Carolina can win is by taking out star players.  Let's see:

Recchi took out Talinder's arm (Buffalo),

and some guy took out Koivu's eye (Canadiens).

And then to put some salt on the wound, the Oilers had a 3-0 lead.  A 3-0 lead!!!  One would think it'd be over by then...nope.  Carolina scores 4 unanswered, Edmonton ties, and then Carolina scores a dinky goal against the backup goaltender.

What a night...dirty *******...

:cursing: NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!   :furious:
 :( Roloson :stretcher:  is not ok. [url]http://www.cbc.ca/sports/hockey/stanleycup2006/storyview.html?/story/stanleycup2006/national/2006/06/06/nhl-oilers-roloson.html[/url]

Sigh...that's kind of what happens when your bones break into pieces...you don't play :(.

I'd hate to admit this...but Roloson was the team.  It's pretty evident that it's over.  While I'm hoping for some miracle to happen, losing Roloson was huge.  Alas, another year passes by in Canada's Stanley Cup drought...

Lol...I'm hoping the Leafs will take us out of our drought.  They're my boys...WOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO LEAFFFFS!!!!!!!!!!!!

Latest and gentlemen, Its OVER!!!!!!!!!!!!!!!!!!!!!!! The Carolina Hurricans have won the Stanley Cup. Conn Smythe goes to the unstoppable Cam Ward, who just plain robbed the Oilers of all except one goal tonight and the Canadian teams continue their drought since 1993. But 2nd year in a row in the finals aint bad.


F(x) = 3-4x/1+x^2 Plot this, then use to solve graphically -F(x)=F(Ax) for different x. Why precisely one solution for x > 3/4 ? Show x must be at least root6 - 2 for there to be a solution for A.

I have drawn the graph and verified it is correct using my graphics calculator. For the -F(x)=F(Ax) part I figured it was F(x) reflected in the x-axis and then see what happens when A varies i.e. A>1, A=1, A<1 ? This may be completely wrong.

Any suggestions?

JasonSullivan,

in its horizontal style, this is typed in wrong.  You need
grouping symbols.  Did you intend:

F(x) = (3 - 4x)/(1 + x^2)?

Or, did you intend:

F(X) = 3 - 4x/(1 + x^2)?

As is, what you have is equivalent to:

F(x) = 3 - 4x + x^2.


geometry
3D geometry
videos
abstract algebra

Do you know any really complex, innovative and playable computer strategy/RPG games worth playing? When I was a bit younger I used to play quite a lot and I think I've played every major title there is, from Civilization through Heroes of M&M to Planescape:Torment. Now most games seem to be boring (I got older :) ), they hardly ever introduce any genuinely new elements into gameplay, and that's especially true about mass-market games, so I'm looking for some less known titles. As for RPG I tried roguelike stuff like ADOM, but that's just hack'n'slash, and I tried Stars!, but I felt there's more bureaucracy there than real gaming. I'd like to find something really complex and difficult, but Google/Wikipedia bring nothing of that kind. Any suggestions?

Oblivion is an AMAZING RPG. WAAAY better than all the MMO crap you see. It's super fun on the Xbox 360, and on the PC you can find some beast mods (I reccomend getting a cheap, wired 360 controller on eBay and hooking it up to your PC with the USB drive, and then playing Oblvion with it).

Strategy games are always more fun on PCs though. Warcraft 3 was beast, but now its old... Starcraft 2 is coming out, however, and it will pwn.

I don't play any of these games any more, though, since they are so awesome that your time gets sucked in a black hole.

Blizzard's RTSs are definitely worth playing, although you probably have played them. Both Warcraft 3 and Starcraft are good.
Diablo 2 seems nice 2. Although they announce Diablo III just recently.

A note about Starcraft and Warcraft, Warcraft focuses much more on micro (good units control). For PRGs, I recommend Paper Mario: Thousand Year Door. Certainly very innovative, and humorous characters make it even better. Also, Super Paper Mario is a good title, probably even more innovative (it's a kind of Platformer/RPG hybrid). However, whether you enjoy Super Paper Mario isnt just a question of whether you like RPGs.

About Oblivion, this isn't my personal opinion seeing as I haven't even played it, but my brother says Oblivion isn't as good as people say it is unless you mod it like crap.

Also, don't be too sure about Starcraft 2 pwning. Remember brawl? Yeah, you get what I mean. However, the attitude of Starcraft people is much different, so I doubt it will be crappy. (Attitude of the people who made brawl: hmm, wavedashing and l-canceling and the other stuff had a huge roll in enriching the metagame of melee... OH, I know, let's take it all out!)

Yeah. Oblivion pwns (ut not as much as Morrowind). I get Game Informer and they take a readers top 5 games ever every week so I will do one:

1. Elder scrolls III: Morrowind, Xbox (RPG)

2. Starcraft PC (RPG/strategy)

3. Super Mario 64 N64 (erm... adventure/rpg?) and its good on the ds.

4. Elder scrolls IV: Oblivion, Xbox360 (RPG)

5 (tied). Spyro (lol I havent played it in years XD): Year of the Dragon Playstation1 (adventure/rpg ish) (all the other games have gone downhill from here...)

5 (tied). Legend of Zelda: Ocarina of Time (I believe the only perfect 10 on IGN other that GTA:IV) Game cube  (I played it on this, not as it originally was on the N64) (RPG)

Yeah, PRGs are really cool in my opinion.

Sidenotes: I own Super Paper Mario for Wii, and I just cant get over the stupid characters. They're all so.... bright. And really dinky. The game is decent but I stopped playing eventually because I find them so annoying.

Other waaay overrated games: Super Mario Galaxy, Pokemon, and Zelda games (although I do like them).

EDIT: I forgot! WoW!!! How could I???? (this goes in the overrated section)

The Honorable Mention: Mario Tennis 64. The only sports game I have ever liked, truly. This deserves to be up top but I forgot it so its honorable mention.

[quote="Anaxerzia"]Morrowind was super fun, though I suggest Oblivion because the graphics are so awesome.[/quote]

Of course, I you want really awesome graphics, you need to make sure your computer is good enough.

About Super Paper Mario, yeah, definitely not quite as good Thousand Year Door, but he was asking for innovative games, and Super Paper Mario certainly falls into this category. What other Platformer/RPG hybrids are there out there?

Dude, legend of zelda is awesome. Done. Wait, actually not done. In twilight princess, one of the few flaws is the boss fights. I mean, they look pretty awesome, but they're just too plain easy. There, now I'm done.

Super Mario 64 isnt RPG for sure. It's a platformer.

Pokemon is pretty fun, you know. Although I think their creative peak has certainly ended. In generation 2, you can see creative pokemon, like hitmontop and corsola. In generation for, you get repulsive design, and a lot of the pokemons have even mroe evolutions (ie, Yanmega, Magmortar, Lickylicky, etc). Also, since they can't make awesome normal pokemon, they put in a crapload of legendaries (ther are around 40 legendary pokemon now).

Oblivion for the Xbox 360 is one of the best RPG games out there. The gameplay is ridiculously long for an Xbox game, and with the DLC, it's even better. I haven't really played it, but my friends and my biology teacher definately recommended this.

The Halo games are....ok. I really hate the co-op but multiplayer is fun. My sisters english teachers husband (mouthful) won 2nd place in a worldwide Halo tournament. He won some amount in the ten-thousands of dollars.

Anaxerzia is right, Oblivions graphics are way better (at least on the 360) and load times are faster. But then again, for how much data on the entire land of Morrowind for one game to store was amazing, to me at least. (Oblivion is really big too.)

Another great game for the Xbox/360 is Fable. My brother rented it one day when my mom was at work and all day we played that game. Then my mom got home and was furious with my brother for bringing home an "M" for Mature game without her knowing.

I agree that LoZ series is awesome but they are way overrated. I believe there are some other games out there that are just as good but almost no one plays them.

Oh yeah... I really just put RPG because I had no idea what it was :D

Yep. I don't own any Pokemon games anymore but I hated all legendaries, basically.

The Halo campaigns are beast. And no, Halo isn't a RPG/strategy game.

If you're going to get Oblivion I'd definately recommend it for the 360 instead of the computer, because you might have to spec up your computer (graphic cards, sounds, video, etc.) and that might cost a lot. Not to mention the sheer memory of Oblivion on your computer. The regular Xbox 360 holds about 15 GB of memory - easily more than enough for hardcore Oblivion. The Xbox 360 Elite has like a ridiculous amount - a little less than 10x more memory than the regular.

You can also try out Rainbow Six Vegas 2 - it's sort of the RPG type - since you get to build you're own suit and weapons and stats. Unfortunately, the graphics are a little disappointing, and the M game might not be okay for some of you youngsters  :P

http://babarageo.com/flash/ginormo/

or http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212793 for comments, strategy discussions, and sort of walkthroughs.

Neverwinter Nights is my all time favorite RPG. It's pretty much Dungeons&Dragons on a computer. If you get bored with the single player game, go onto the multiplayer online. You can make your own modules for it too.

As of Legend of Zelda, I haven't run into one that I didn't like. Windwaker was my favorite, although it did get a bit tedious once. The bosses in it were awesome though.

And Pokemon, well.....not quite....

[quote="cyberspace"]Windwaker was my favorite, although it did get a bit tedious once. [quote]
Let me guess: it was gathering the triforce.  :P . That was kinda lame.

Oh, another thing, if you're going to get a LoZ game, know this:

Ocarina of Time: Considered BEST EVA. Held in as high reverence as Super Mario 64 (and that's saying a lot).
Majora's Mask: Shorter, but it's still nice. If you like a creepy feel, this is nice, too. Beware though, getting fierce deity mask is like... lame
Wink Waker: Innovative in its graphics, although whether you like it is your opinion (some people think it's beautiful, others think it looks stupid). Sailing might be annoying, and gathering  the triforce WILL be annoying.
Twilight Princess: Boss fights are too easy. Otherwise, this is good in pretty much every way. Very good graphics, too. I mean, the actual image quality is mediocre, but the art design is simply beautiful, and I think everyone agrees on that. You'll be dissappointed my the graphics at first, but it really grows on you.

[quote="cyberspace"]Yes, it was gathering the triforce.... :(

also with regards to LoZ:
DON'T play the Minish Cap it's stupid.
Ocarina of Time was awesome. The final battle was great.
I have yet to play majoras mask
Windwaker as I said was great. The final battle was my favorite of all time!!!
Link to the Past/Four Swords is a fun game if you disregard the terrible graphics.
I still haven't finished Twilight Princess but I was majorly impressed by the graphics.

Metroid Prime was verrry fun and had great graphics, but I suppose it's not quite an RPG/strategy game

All of the Final Fantasys are GREAT especially VII and X. XII is pretty good too, but VII has the best plot and X has the best graphics.

But still, Neverwinter Nights is the best RPG I've ever played.[/quote]

Hm? Why is Minish Cap stupid? I thought it was pretty good...

Also, funny thing is, FFVII and X are generally regarded as the best ones.

[quote="Nerd_of_the_Ages"]
Hm? Why is Minish Cap stupid? I thought it was pretty good...
[/quote]

Well, it was ok, it just didn't live up to my expectations of a LoZ game. Even Link to the Past is wayyy better


induction
combinatorics unsolved
combinatorics

If $n$ lines in plane have 2 points in intersection, prove that they must have at least $n-1$ intersection points.

We will call the intersection points "balls" for clarity.

We prove the following by induction:
[b]Lemma.[/b] Let's have $n$ lines such that:
1)they are not all paralel
2)they do not all pass through single point.
Then there are at least $n-1$ balls, and the only possible exception is in the following cases
i)all except one are paralel and one is not paralel to them
ii)there are 2 paralel lines and the others pass through single point on one of these two lines.

The base n=3 is clear! Three lines either form triangle, or pass through one point, or are paralel, or [b]there are 2 paralel lines and the third is not paralel to them[/b](this are cases i) and  ii) ).

Suppose we have proven the lemma for n=3,..,k.
In the case n=k+1 we simply remove one line L. But we should choose the line L to be such that it passes through a 2-line ball. Such point exist due to Silvester's theorem as grobber says ;)
If the other k lines fail to satisfy conditions 1) or 2) then when we return L we get that the k+1 lines satisfy i) or ii).
If the other k lines satisfy 1) and 2) then there are at least k-1 balls on them. If there are more than k-1 balls on them then it's OK, unless the conditions i) and ii) but if after we add the line L there are exactly k points then we have added one point to configuration of i) or ii) and we will reach again to configuration i) or ii). :)
If there are exactly k-1 balls on them then then satisfy i) or ii). And adding a line to configuration that satisfies i) of ii) will add at least one ball (this is easy to see but lazy to write :D).
So, we've proven the Lemma for n=k+1.

The only thing left to complete the proof is to see that the lines given satisfy 1) and 2)!!!!

P.S. I edited because firstly I missed the induction proof for the conditions i) and ii) but now it is corrected.

You made a huge mistake!
In case of $k + 1$ lines, when you remove one line $L$, and other $k$ lines produce EXACTLY $k$ intersection points, maybe line $L$ can pass only through existing intersection points (not making new ones) and we again have equality!

Here's a proof by induction on $n$. Assume $n\ge 4$.

I'll use the well-known problem (due to Sylvester, I believe, and discussed before) which states that given finitely many lines in the plane, not all parallel, such that for any intersection point of two of them, there is a third passing through it, these lines must all pass through the same point. Check [url=http://www.ams.org/featurecolumn/archive/sylvester1.html]this[/url] out. It's a great article about this problem.

We apply the above-mentioned result to our configuration. Since our lines are neither parallel nor cuncurrent, there must be a point through which only two lines pass. Remove one of these lines (call i $a$). We are now left with $n-1$ lines. If they are parallel, after adding $a$ back we get $\ge n-1$ points. If they're concurrent, again, after adding back $a$, which cannot pass through the concurrence point, we get $\ge n-1$ points. Finally, assume these $n-1$ lines are neither parallel nor concurrent. Then they determine $\ge 2$ points, so, by the induction hypothesis, they determine $\ge n-2$ points. After adding back $a$, we get at least one extra point, the one through which only $a$ and another line pass. We thus have $\ge n-1$ points.


induction
inequalities
inequalities proposed

Prove that for all $ n\geq2$ and $ n\in\mathbb N$ holds:
$ \sqrt[n]{\dbinom{2n+1}{n}}\geq2\left(1+\frac{1}{\sqrt[n]{n+1}}\right).$

$ {f(n)=\left(\frac{(2 n+1)!}{n! (n+1)!}\right)^{1/n}}$
$ {g(n)=1+\frac{1}{(n+1)^{1/n}}}$

$ {f(n+1)\geq f(n)}$
$ {g(n+1)\geq g(n)}$

We just need to proove the first, because the second is trivial.

It's a little bit trouble, but can be showed

$ {f(n)=e+\left(-\frac{1}{2}(3 e)+\frac{e \pi^{2}}{6}\right) n+O\left(n^{2}\right)}$

Or in another words, $ f$ is increasing. $ g$ is increasing too. so the minimum is in 2

$ {\frac{f(2)}{g(2)}=\frac{\sqrt{10}}{1+\frac{1}{\sqrt{3}}}\geq 2}$

i dont know a good way to proove that f is increasing, but since it is prooved, it's done

that kind of doesn't work
you never proved that $ \frac{f}{g}$ is increasing
your way could have proved that $ \frac{g}{f}\geq \frac{1}{2}$

Using induction we only have to prove that:
\[ C_{2n+3}^{n+1}\geq 2^{n+1}\left( 1+\frac{1}{\sqrt[n+1]{n+2}}\right)^{n+1}\]
But
\[ \left( 1+\frac{1}{\sqrt[n]{n+1}}\right)^{n+1}\geq \left( 1+\frac{1}{\sqrt[n+1]{n+2}}\right)^{n+1}\]
therefore we only have to prove that:
\[ C_{2n+3}^{n+1}\geq 2^{n+1}\left( 1+\frac{1}{\sqrt[n]{n+1}}\right)^{n+1}\]

\[ \Longleftrightarrow \]

\[ \frac{(2n+2)(2n+3)}{(n+1)(n+2)}\cdot C_{2n+1}^{n}\geq 2\cdot 2^{n}\left( 1+\frac{1}{\sqrt[n]{n+1}}\right)^{n+1}\]
Using the induction hypothesys it remains to prove that:
\[ \frac{(2n+2)(2n+3)}{(n+1)(n+2)}\geq 2\left(1+\frac{1}{\sqrt[n]{n+1}}\right) \]
which turns out to be true after some computations.(We may use Bernoilli's inequality as well)

[quote="Svejk"]
But
\[ \left( 1+\frac{1}{\sqrt[n]{n+1}}\right)^{n+1}\geq \left( 1+\frac{1}{\sqrt[n+1]{n+2}}\right)^{n+1}\]
[/quote]
I think, it's wrong, Svejk :wink:


The tree multiplies buy 1.25
So 1.25*10=12.5 feet, which is 150 in.

[quote="asianpower"]Can somebody help me with this problem?

Bob is standing next to a tree that is 40 feet tall. The shadow of the tree is 50 feet. Bob is 10 feet tall. How long is his shadow in inches?[/quote]

Set up a proportion:

$\frac{40}{50}= \frac{10}{x}$.

Cross multiply:

$500=40x$

$x=12.5$

12.5 feet is 150 inches.

[hide="answer to proportion prob"] We can set up the ratios
$\frac{40}{50}=\frac{10}{x}$
$40x=500$
$4x=50$
$x=12.5$
But the question asked for inches, so
$12.5*12=25*6=150$
So our answer is [b]150[/b] inches.[/hide]

[hide]
$\frac{40}{50}=\frac{10}{x}$
After simplifying
$x=\frac{25}{2}$
then
multiply by $12$
This equals
$25*6=\fbox{150}$[/hide]

I'm not sure, but I suppose:

[hide]150 inches[/hide]  :oops:


I recently ran into a probability problem in AoPS Volume 2, and reduced it to finding the area of one of the sections of the following figure:
[asy]size(100); draw(arc((0,0),1,0,90)); draw(arc((1,0),1,90,180)); draw(arc((1,1),1,180,270)); draw(arc((0,1),1,270,360)); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);[/asy]
(four overlapping quarter-circles inscribed in a unit square).

When I looked up the answer in the solution manual, it referenced ways to find areas of overlap among circles that had been taught in Volume 1.  Does anyone know what those methods are?

maybe you could try it yourself.
[hide="hint"]
[asy]size(100); path A, B, C, D; draw(arc((0,0),1,0,90)); draw(arc((1,0),1,90,180)); draw(arc((1,1),1,180,270)); draw(arc((0,1),1,270,360)); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); A = arc((0,0),1,0,90); B = arc((1,0),1,90,180); C = arc((1,1),1,180,270); D = arc((0,1),1,270,360); dot(intersectionpoint(A,B)); dot(intersectionpoint(A,D)); dot((1,0)^^(0,1)); draw(intersectionpoint(A,B)--intersectionpoint(B,C),linetype("8 8")); draw(intersectionpoint(B,C)--intersectionpoint(C,D),linetype("8 8")); draw(intersectionpoint(C,D)--intersectionpoint(D,A),linetype("8 8")); draw(intersectionpoint(D,A)--intersectionpoint(A,B),linetype("8 8")); draw(intersectionpoint(D,A)--(1,0),linetype("8 8")); draw(intersectionpoint(D,C)--(1,0),linetype("8 8")); draw(intersectionpoint(B,A)--(0,1),linetype("8 8")); draw(intersectionpoint(B,C)--(0,1),linetype("8 8"));[/asy]
prove that the four points are equidistant on the circle
[/hide]

I think I know what to do from here.  The area that I was searching for was the area outside the "lemons", i.e., the area that is inside only two of the circles.

All of the following calculations are of areas within the unit square.

The circles would have to intersect where either the x- or y-value is 1/2, which is at $30^\circ$ and $60^\circ$.  To find this, I could first find the area of the central region, which consists of a square with side length $\sqrt{2(\frac{\sqrt3}2-\frac12)^{2}}=\sqrt{2(1-\frac{\sqrt3}2)}$ or an area of $2-\sqrt3$ as well as four $30^\circ$ arcs cut off by their respective secants, each of which would have an area of $\frac\pi6-\frac12\sin{30^\circ}=\frac\pi6-\frac14$ for a total area of $\frac{2\pi}3+1-\sqrt3$.

Now, the area outside a given circle is $1-\frac\pi2$, so each "lemon" must have an area of $1-2(1-\frac\pi2)=\pi-1$.  Subtracting from this the area of the central region and dividing by two, we find that each "bent triangle" must have an area of $\frac{\frac\pi3-2+\sqrt3}2=\frac\pi6-1+\frac{\sqrt3}2$.

Multiplying by four and adding the area of the center, we have $\frac{4\pi}3-3+\sqrt3$ for the area that is the intersection of at least three circles.  Subtracting from 1, we find the desired area is $4-\frac{4\pi}3-\sqrt3$.

The book's answer is $4-\frac{2\pi}3-\sqrt3$, but I can't find any holes in my work.  Am I missing something here, or did the AoPS folks make a mistake on this one?

By the way, thanks everyone for helping me understand what they were doing.

[quote="roadnottaken"]we find the desired area is $4-\frac{4\pi}3-\sqrt3$.
[/quote]
How can the area be a negative number?

i think your error is here: :wink: [quote="roadnottaken"]Now, the area outside a given circle is $1-\frac\pi2$[/quote]

OK, I see what I did.  I was thinking in terms of the radian measure rather than the area.  Changing from $\frac\pi2$ to $\frac\pi4$ does indeed clear everything up (and I really need to work on careless errors).  Thanks again!


linear algebra
matrix
search
linear algebra unsolved

Let's define Jacobi matrix $\begin{displaymath}\mathbf{J}=\left( \begin{array}{cccccc}{a_1} & {-b_1} & 0 & \ldots & 0 & 0 \\ {-c_1}& a_2 & {-b_2} & \ldots & 0 & 0 \\ 0 & {-c_2} & a_3 & \ldots & 0 & 0 \\ \ldots & \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & 0 & \ldots &{-c_{n-1}} & {a_n} \end{array} \right) \end{displaymath}$, $a_i,b_i,c_i\in\mathbb{R}$,$b_ic_i>0,1\leq i \leq{n-1}$
Prove that the $Spec(J)$ is always real and simple(it means that all the eigenvalues of $J$ has one algebraic multiplicity)

just a try:
we can search $D=diag(d_1,d_2,..,d_n)$ such that $S=D^{-1}jD$ is symetric.
Then a few calculation show that:
$\frac{d_2}{d_1}=\sqrt{\frac{c_1}{b_1}}$
$\frac{d_3}{d_2}=\sqrt{\frac{c_2}{b_2}}$
..
$\frac{d_{n}}{d_{n-1}}=\sqrt{\frac{c_{n-1}}{b_{n-1}}}$
So we can find $D$ with arbitrary $d_1$.
Then $J$ is similar to a symetric matrix : its spectrum is real.
(I did not really check the computations ..)


No esta escrita exacta:

Empezando con $ 1$, encuentra una secuencia de operaciones de $ A$ y $ B$ para llegar a $ \dfrac{2009}{2^{2009}}$

Operación $ A$: Divide el numero entre dos.
Operación $ B$: Resta el numero al numero $ 1$.

Ejemplos:
Si empieza con $ \dfrac{5}{8}$ y haces la operación $ A$, el resultado es $ \dfrac{5}{8}\cdot \dfrac{1}{2}\equal{}\dfrac{5}{16}$.

Si empieza con $ \dfrac{1}{16}$ y haces la operación $ B$, el resultado es $ 1\minus{}\dfrac{1}{16}\equal{}\dfrac{15}{16}$.

[hide="mi respuesta"]
$ 3A,B,2A,B,A,B,5A,B,1998A$
[/hide]

You want me post this in the Pre-Olympiad/Olympiad section in English?

if you want, also if you remember the other problems you should post them too, I think these should go in the intermediate level.

esa fue mi solución tambien, tu participaste en la olimpiada??

Si y tambien participo modularmarc101

Si tienes los problemas por favor ponlas en el foro, ya yo puse la primera y tercera pregunta en el foro intermedio en ingles.


I have another question from 2004's National Target Round:

A collection S of integers is defined by the following 3 rules: (i) 2 is in S; (II) for every x in S, 3x and x+7 also exist in S; (III) no integers except those defined by the first and second rules are in S.  What is the smallest integer greater than 2004 which is not in S?

If all you are going to do is copy and paste the solution from the 2004 Solutions, it sometimes doesn't help, so please send some of your [size=200][u][i][b]own [/b][/i][/u][/size] speculations and explanations as well. :thumbup:

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=789955&search_id=1699847139#789955]rcv's excellent explanation here.[/url] The Search button is your friend.


Can someone teach me how to solve the following questions?I don't know how to solve it

Show that
a) if u is orthogonal to both v and w, then u is orthogonal to kv + lw for all scalars k and l.
b) if a, b, c, and d lie in the same plane, then ( a × b ) × ( c × d ) = 0.

Mhmm for a), since u is orthogonal to v, we can say $ <u,v> \equal{} u \cdot v \equal{} 0$. Similarly with w. (I did the dot product assuming thats what you've learned).
Now $ <u,\alpha v> \equal{} \alpha<u,v> \equal{} \alpha 0 \equal{} 0$ Do the same with the other vector and add them.

For b) do you know what the cross product of two parallel vectors are?

I don't know !! :(  :( I just newbie on this subject

Okay, do you understand what the cross product is then? If so how have you learned to calculate it?

Also, for admin, could you move this? It isn't an Olympiad question :)


1. (Romanian Olympiad -- i checked this site but i didnt find a solution) Let ABC be a triangle with angle BAC = 60. Let
P be a point in its interior so that PA=1,PB=2 and PC=3. Find the
maximum area of the triangle.

2. (HK Olympiad) of all the quadrilaterals of area 1, find the
minimum possible value for the sum of sides and diagonals.

Ok I think what you're looking for is the minimum of the sum , which is indeed 4 + 2*sqrt(2).

Let a,b,c,d be the sides (in this order) and e,f the diagonals .

ab*sin(something) + cd*sin(something) = 2*S , so ab + cd>=2S=2
ad*sin(something) + bc*sin(something) = 2*s , so ad + bc>=2S=2
ef * sin(something) = 2*S , so ef >= 2

(a+c)(b+d) >= 4 , so a+b+c+d >= 4 , by AM-GM
ef >= 2 , so e+f >= 2*sqrt(2) by AM-GM and we are done.

If you want the maximum sum , I think there is no maximum. Here is why : push the quadrilateral to the limit such that it is degenerate , a triangle , like this : B is the middle of AC. We get a triangle , and we can make the AC side as long as we want , while keeping the others > 0. Even if you're asking for the maximum for a convex quadrilateral , you can "move" the points a little once you've achieved an arbitrarily large sum of sides + diagonals and still get that there is no maximum . I have no time to explain more , but I think it's pretty clear what I mean.

thanks for the reply.  and you are right, it is minimum :S

i understand that there is a formula that says ef*sin(t)=2S, where t is the angle between the diagonals.  do you know how to prove this little fact?

That's really easy:

You calculate the area as being the sum between the areas of the 4 smaller triangles formed by the point of intersection of the diagonals and 2 consecutive vertices. Check it out, keeping in mind that sinx=sin(pi-x).

yes, i should've seen that.  thank you grobber

for the 1st problem just check the problem from the Geometry cathegory, which has the following topic: maximize area

cheers! :D


search
inequalities
algebra
polynomial
number theory proposed
number theory

suppose that $ x_{1},x_{2},\dots,x_{n}$ are integer numbers.prove that $ \prod_{1\leq i,j\leq n} \frac {x_{i} \minus{} x_{j}}{i \minus{} j}$ is integer number.

See Albanian Eagle's solutions [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1007862998&t=290405]here[/url].

There have been many proofs given to this on this forum. Here is yet another one.
If we denote $ \lambda$ to be the partition with rows of length $ a_i\plus{}i$ then
\[ \prod_{1\le i<j\le n}\frac{a_i\minus{}a_j}{i\minus{}j}\equal{}s_{\lambda}(1,1,...,1)\]
where $ s_{\lambda}$ denotes the Schur polynomial corresponding to $ \lambda$.
The result follows once we know that the Schur polynomials have integer coefficients (Obvious from either of the Giambelli's formulas).


modular arithmetic
function
number theory
relatively prime

Find the remainder when $ 7^{7^8}$ is divided by $ 10000$.

[hide="outline"]First find $ \phi{10,000}$. Then compute $ 7^8\pmod{\phi{10000}}$. Then express this in binary and use consecutive squaring.[/hide]

I will continue my advocacy of the [url=http://en.wikipedia.org/wiki/Carmichael_function]Carmichael function[/url] over the totient function.  At higher powers of $ 10$, it really starts to help.

Are you sure it's $ \mod 10000$? Doing this $ \mod 10$ or even $ \mod 100$ is not that big a deal, but anything more than that is just plain annoying.

Anyway, $ \lambda(10000)\equal{}500$, so you have to find $ 7^8\bmod 500$, then find $ 7$ raised to this power $ \mod 10000$.

Because Carmichael's lambda function $ \lambda(n)$ guarantees to give you the smallest $ k$ such that $ a^k\equiv 1\bmod n$ for all $ a\in\mathbb{Z}$ such that $ (a,n)\equal{}1$.

Think about why you'd want to find $ 7^8\bmod 500$. What significance would that have?

[quote="ThinkFlow"]Why does that process work? :maybe:[/quote]
The Carmichael function [i]almost[/i] gives you the same result that you would get from the Chinese Remainder Theorem applied to the prime powers of the modulus, except that $ \lambda(2^n) \equal{} 2^{n\minus{}2}$ for $ n \ge 3$ instead of $ 2^{n\minus{}1}$.  This is because there are no primitive roots $ \bmod 2^n$ for $ n \ge 3$, which is somewhat subtle.  You can verify this for small moduli for yourself:  for example, you should be aware that $ x^2 \equiv 0, 1, 4 \bmod 8$, so in other words for $ x$ relatively prime to $ 8$ we have $ x^2 \equiv 1 \bmod 8$ rather than $ \phi(8) \equal{} 4$.


Source : IMLEL Meet #4

Davey joined Abby, Billy, and Casey outside to play in the snow. They all wore
a hat and a scarf.  How many ways can all four kids exchange hats and scarves so
that nobody has their own items and nobody has two items from the same person?

[hide]There are 4 choices for the first person.  He then has 3 choices for a hat and 2 choices for a scarf.  There are 3 choices for the next person, 2 choices for a hat and 1 choice for the scarf.  There are only 2 more ways the last two people can put on the remaining clothes, for a total of $4\times3\times2\times3\times2\times1\times2=288$ combinations.[/hide]
However, this reasoning fails because of several things.  First, the order in which the people recieve the clothes does not matter.  Second, I counted the number of ways to give out the scarves and hats incorrectly.  I am sorry about that, and I have a new answer.
[hide="NEW ANSWER"]Call the people A, B, C, and D, and their items, hat and scarf A, hat and scarf B, hat and scarf D, and hat and scarf D, respectively.  Note:  Person A is the first person to get something, person B is the second person...  Person A is given one of the hats B, C, or D.  This can be done in 3 ways.  Next, person B (person A has his hat) has 3 choices for a hat.  The remaining people are C and D.  If person B picked person A's hat, then there is only one way to give the hats to people C and D.  If person B did not get person A's hat, there still is only one way to pass out the remaining 2 hats.  This makes a total of $3\times3=9$ ways to pass out the hats.

For the scarves, there are only 2 ways to give them out, where person A has B's hat, person B has C's hat...
C, A, D, B
D, A, B, C

This gives a total of $9\times2=18$ ways to pass out the articles of clothing.[/hide]

[hide="i got"]3*3*2*1=18. Square it to get 324. Subract 4 to get 320[/hide]

So I was reading how it was better to do more challenging problems and do less chatting about scores and rankings.  Well here is your challenge.  So far the two answers given are pretty far off.  Anyone up to the challenge?  This problem came from a local math competition and it was a team round problem.  Good luck and keep working at it.

If you know a quicker way, or a simpler explanation of the solution, could you please explain?
[hide]
Person A has three possible hats to pick, B, C, and D.

The person who's hat is picked up by person A also has three choices.

No matter what choice the second person takes, one of the two remaining people has only one hat left to take, because they cannot pick their own, and two hats of other people have been chosen.

Therefore, there are [b]3*3=9 [/b]possible hat arrangements.

Now, each person is left with two possible picks of scarves. Each individual scarf has the oppurtunity to be chosen by two people.

When person A takes a scarf, there is another person who also could have chosen that scarf, but will have to take a different one, and the other person who could have chosen that scarf will also have to take their second choice.

Therefore, there are [b]2[/b] possible hat arrangements. [b]2*9=18   [/b] total possibilities.
[/hide]

I agree with azndemon43.

BTW there was a mathcounts warmup just like this problem:
Four people are sitting around a table. In how many ways can their nametags be passed out so that everyone gets the wrong nametag?

[hide]The first nametag you have 3 places to give it out and satisfy the problem.
The second nametag you have 2 places to give it out and satisfy the problem.
The third and fourth nametags you split to the people who do not have that name.  3*2=6?[/hide]

is the answer 27,  I am stupid enough to try to count all of them...

yeah, there are 9 arrangements that don't have anything in the original position, but I think I counted 27 that matchup without two things in the same row...

random guess:

[hide]22? i counted the number for hats... and got 11, and just multipled that by 2... i don't know why[/hide]

everyone else has already tried my way so here's my [hide="answer"] 280 ways. if you have one person who has anyone's hat and scarf, then there will be two people with 4 choices on what to get and one with three.  I don't think it works, though, but it's different from everyone else's answers. :| [/hide]

Frost13, did anybody get it right?

the answer to the question I posted is
[hide]
9--same logic as chosing the hats
write out the combos to see why 6 doesn't work
hint:
[hide]
If you have to match up the people ABCD to their tags abcd and you have three choices of where to put a, and you put it on B, then for b you have 3 choices still...
[/hide]
[/hide]

The first time I saw the title of the topic I thought that you guys that were going to nationals wanted to trade clothes like you were trading pins and i started :rotfl:

[quote="frost13"][quote="binonunquineist"][quote="frost13"]binonunquineist isn't (4*4-4)*4 = 48???????????

What is the reasoning for your method?

I'll give it a little longer and then post the solution for the actual answer :)[/quote]

oops, typo, i meant (4*4-4)*[b]2[/b] = 24.  :blush:

o well, its wrong anyways.[/quote]

[hide]Well, it's not wrong, 24 is the answer.  I was just trying to see why you think that would give you your answer.  Can you share your logic?  Here is the solution and correct answer :

From Abby’s point of view, there are three different choices for the hat she can
wear. Once she has chosen a hat, there are two remaining choices for the scarf she
can wear. Thus there are 3 x 2 = 6 ways Abby can pick a hat and a scarf. That’s
the easy part. Now we have to consider what the other three kids can do after
Abby makes her selection. Suppose Abby chooses Billy’s hat and Casey’s scarf
(ABC). Notice that Davey still has his own hat and his own scarf.
If Davey gives his hat to Billy, then everything else is forced, and this is the result:
(1) ABC, BDA, CAD, DCB. If instead Billy gives his hat to Casey, then there are
three more ways the hats and scarves can be exchanged. They are
(2) ABC, BAD, CDB, DCA; (3) ABC, BAD, CDA, DCB; and (4) ABC, BCD,
CDA, DAB.
Abby’s original selection was one of six possible choices. Each of these six
choices for Abby gives the other kids four possible ways to exchange hats and
scarves. Thus there are 6 x 4 = 24 ways the four kids can exchange hats and
scarves so that nobody has their own items and nobody has two items from the
same person.

Hope that makes sense.[/hide][/quote]

Yay!!! So I win. Here's what I did, although its not the best method, it still comes to the correct answer.

Each person can get 4 hats, since there are 4 people its 4*4
4 of these ways match someone to their own hat, so subtract 4
For scarves, you can't give someone their own scarf and you can't give anyone the scarf owned by the same person as their hat, so you can give everyone one of 2 scarves. So multiply by 2:
(4*4-4)2 = 24.

It would be quite embarassing if it turns out that frost13 is lying.


Bekah has four brass house number digits: 2, 3, 5 and 7. How
many distinct numbers can she form using one or more of the
digits?

4 ways for 1 digit
4*3 for 2 digits
4*3*2 for 3 digits
4*3*2*1 for 4 digits
4+12+24+24=64.

Note: Fantasy, it says 1 or more digits.


AMC
AIME
ARML
Stanford
college
mathleague.org
Stanford Math Tournament

Hi all

I go to a small private high school (Crystal Springs), and we don't have a very large math community here (I'm the only consistent AIME qualifier left at my school).  What math contests are there that I can do without needing other people from my school?  I do AMC/AIME/etc., BAMO, and I try to get enough people to do the NAT series.  I'm also going to do ARML this year.  What other contests/competitions can I do without needing a team from my school?  Also, are there any good science competitions for which I won't need a school team?  For what it's worth, I live on Stanford campus and I'm willing to travel for them if I have to.

Thanks
Sam

You can definitely do the mathleague.org series--check the website for the schedule later--and you should definitely do this if you're planning on joining the ARML team, as it serves as a team tryout.  SMT (Stanford Math Tournament) in February will be happy to have you compete, and will even set you up with an ad hoc team if you want.

Sam, I'm fairly sure I know you...  You're always welcome to come to Menlo for things that need a larger team (and bring Nathan Tsai with you xD)... We've got a fairly large group of people who are pretty consistently good, and you'd have a lot of fun. If you're interested, leave me a private message on here and I'll give you some contact information.


[quote="kimby_102"][quote="edyfox"]Given $a_{i}\in\mathbb{N}(1\leq i \leq n)$ and $s=a_{1}+a_{2}+...+a_{n}$. Show that $(1+a_{1})(1+a_{2})...(1+a_{n})<1+s+\frac{s^{2}}{2!}+...+\frac{s^{n}}{n!}$ for all positive number $n>1$.[/quote]
we just apply AM-GM inequality.
$(1+a_{1})(1+a_{2})\cdot\cdot\cdot(1+a_{n})\leq \left(1+\frac{s}{n}\right)^{n}$.
now use binomial theorem to get the result.[/quote]To complete kimby_102's solution, the inequality is strict due to the following lemma:

[b]Lemma:[/b] For $n\ge k>1$, $\frac{n!}{(n-k)!}<n^{k}$
Proof? You do it :P

Using this lemma, we get $\binom{n}{k}\frac1{n^{k}}<\frac1{k!}$ and the result follows.


calculus
integration
parameterization
trigonometry
logarithms
calculus computations

Ok, I found the solution.

[b]Lemma:[/b] For $ |a|\plus{}|b|>0$ we have: $ I(a,b)\equal{}\int_0^{\frac{\pi}{2}}\log(a^2\sin^2{x}\plus{}b^2\cos^2{x})dx\equal{}\pi\log{|a|\plus{}|b|}\minus{}\pi\log{2}$.

[b][i]Proof:[/i][/b]

We differentiate w.r.t. $ a$ or $ b$ and then integrate w.r.t. $ a$ or $ b$.

[b]For the problem:[/b]

We just rewrite the argument $ 1\minus{}2a\cos{x}\plus{}a^{2}$ in the form $ (1\minus{}a)^2 \cos^2{\frac{x}{2}}\plus{}(1\plus{}a)^2\sin^2{\frac{x}{2}}.$ After substitution $ t\equal{}\frac{x}{2}$ and the use of upper lemma we obtain $ 2\pi\log{a}$, if $ a\neq0$ and $ 0$ otherwise.

There is another approach:

$ a = 0$ trivial, so assume $ a \neq 0$

It is to note that $ I(a) = \int_{0}^{\pi} \log{(a - e^{ix})(a - e^{ - ix}) dx = \int_{0}^{\pi} (\log{(a - e^{ix})} + \log({a - e^{ - ix}}))dx}$

Making ths substitution in the second integral of $ u = - x$ we get $ du = - dx$ and the second integral becomes:

$ - \int_{0}^{ - \pi} (\log{(a - e^{iu})}du = \int_{ - \pi}^{0} (\log{(a - e^{ix})} du$

Combine this integral with the first, making the substitution $ z = a - e^{ix}$ such that $ dz = - i(a - z)dx$. We now have a complete and closed contour, $ \int_{\gamma} \frac {i\log{(z)}}{a - z} dz$. So the residue at $ z = a$ is $ - i\log{a}$, since the paramaterization of $ \gamma$ involves clockwise circulation. Multiplying by $ 2\pi i$ (Cauchy Residue Theorem) will therefore give $ I(a) = 2\pi \log{a}$

Haha. I just completely ignored jumping over a branch cut...

One of the earliest problems I solved on this site - we're talking 2004 or 2005 - was this one. Perhaps kunny posted it? I've been look for that topic but haven't found it yet.

Thanks, Null. To copy over the solution from that thread,

$ I(a) \equal{} \int^{\pi}_{0}\ln(1 \minus{} 2a\cos{x} \plus{} a^2)dx\equal{}\begin{cases}2\pi\ln|a|&\text{for }|a|>1\\0&\text{for }|a|\le 1\end{cases}.$

For further explanation, follow Null's link.

I see, where I made a mistake in my second post. I should have taken into account three possibilities, that is:

$ |1-a|+|1+a|=\{\begin{array}{ll}
 2&; |a|\leq 1\\
2a&; a>1\\
-2a&; a<-1
\end{array}
\;$

Then we get:

$ 2\pi(\log{(|1-a|+|1+a|)}-\log{2})=\{\begin{array}{ll}
 0&;|a|\leq 1\\
2\pi\log{a}&; a>1\\
2\pi\log{(-a)}&; a<-1
\end{array}
\;$

which can be squeezed to:

$ 2\pi(\log{(|1-a|+|1+a|)}-\log{2})=
\{\begin{array}{ll}
 0&;|a|\leq 1\\
2\pi\log{|a|}&; |a|>1
\end{array}\;$

My argument actually works perfectly. I realise that my contour does not encompass the branch point of $ \log$ for $ |a| > 1$...

But, I can't really believe no one has seen this already..:

For $ |a| < 1$

Rewrite the original integral as:

$ \int_{0}^{\pi} \log{\left(a^2\left(\frac {1}{a^2} \minus{} \frac {2}{a} \cos{x} \plus{} 1 \right) \right)} dx$.

$ \equal{} \int_{0}^{\pi} \left(2\log{|a|} \plus{} \log{\left(\frac {1}{a^2} \minus{} \frac {2}{a} \cos{x} \plus{} 1 \right)}\right) dx$

So we can trivially do the first integral: $ 2\pi \log{|a|}$

The second integral is just the original replaced with $ \frac {1}{|a|} > 1$, and so calculates to $ 2 \pi \log{\frac {1}{|a|}}$

When these are added together, we get the $ 0$ result!


2k symbols 1,1',2,2',.....k,k'
number of orderings so no j,j' next to each other?

i think its easy with inclusion exclusion, reminds me of the problem of seating married couples around a circular table.  just keep your notation clean and its probably straightforward.

See http://oeis.org/A114938 for the formula and related stuff.


geometry
rectangle
rhombus
geometry unsolved

A convex quadrangle $ ABCD$ is inscribed in a circle with center $ O$. The angles $ AOB, BOC, COD$ and $ DOA$, taken in some order, are of the same size as the angles of the quadrangle $ ABCD$.
Prove that $ ABCD$ is a square

Because $ ABCD$ is cyclic, $ \angle{BCD}+\angle{BAD}= 180^{o}$, and then a pair of angles taken from the set $ \angle{BOA},\angle{BOC},\angle{COD},\angle{DOA}$ must add up to $ 180^{o}$. Then as all these four angles together form a $ 360^{o}$ angle and have the same vertex we get 2 cases:
1. The two angles share a ray, wolog $ \angle{BOA}+\angle{BOC}= 180^{o}$. But then, $ O$ lies on $ AC$ so $ AC$ is a diameter hence $ \angle{ABC}=\angle{CDA}= 90^{o}$. then, again by the condition of the problem we must have one of angles $ \angle{BOA},\angle{BOC},\angle{COD},\angle{DOA}$ equal to $ 90^{o}$ so wolog $ \angle{COD}= 90^{o}$ then $ \angle{AOD}= 90^{o}$ so $ CD = AD$ and $ \angle{OAD}=\angle{OCD}= 45^{o}$. Then if $ \angle{BCO}= x$ then $ \angle{BAC}= 90^{o}-x$ as $ \angle{ABC}= 90^{o}$, then we get $ \angle{BAD}= 135^{o}-x$ and $ \angle{BCD}= 45^{o}+x$ while $ \angle{BOC}= 180^{o}-2x$ and $ \angle{BOA}= 2x$. Then we get $ 2x = 45^{o}+x$ giving $ x = 45^{o}$ or $ 2x = 135^{o}-x$ also giving $ x = 45^{o}$. Then $ \angle{BCD}=\angle{BAD}= 90^{o}$ therefore $ \angle{BOC}=\angle{BOA}= 90^{o}$ so $ BC = CD = AD = BA$ as triangles $ BOC, COD, DOA, AOB$ are congruent so as all angles of $ ABCD$ are $ 90^{o}$ $ ABCD$ is a square.
2. The two angles do not share a ray, wolog $ \angle{BOA}+\angle{COD}= 180^{o}$ then $ \angle{BDA}+\angle{CAD}= 90^{o}$ therefore $ BD$ and $ AC$ intersect at a $ 90^{o}$ angle giving $ ABCD$ is an isosceles trapezoid. Therefore, wolog $ BA = CD$ and then $ BC$ is parallel to $ AD$, but then $ \angle{ABC}=\angle{BCD}$ and $ \angle{BAD}=\angle{CDA}$. If all four angles are equal we get $ ABCD$ as a rectangle and complete the prff in the same way as in case 1. Otherwise, by the condition of the problem, this means that either $ \angle{BOA}=\angle{COD}$ and $ \angle{BOC}=\angle{AOD}$ resulting in each of the four angles being $ 90^{o}$ as $ \angle{BOA}+\angle{COD}= 180^{o}$ or $ \angle{BOA}=\angle{BOC}$ and $ \angle{AOD}=\angle{COD}$. But then $ AB = BC$ and $ AD = CD$ so $ AB = BC = CD = DA$ so $ ABCD$ is a rhombus and because it is inscribed in a circle, it is a square.
QED


trigonometry
geometry
circumcircle
geometric transformation
reflection
angle bisector
perpendicular bisector

Let $A$ and $B$ be two fixed points on a fixed circle $S$, and let $v$ be a fixed length. Two variable points $C$ and $D$ lie on the circle $S$ such that $CD=v$ and the quadrilateral $ABCD$ is convex. Two points $X$ and $Y$ are chosen on the rays $AC$ and $BC$ such that $AX=AD$ and $BY=BD$. Prove that the distance $XY$ remains constant while the points $C$ and $D$ vary.

This problem is not quite correct. See my post below.

The quadrangle $XYCD$ is cyclic. This is follows from angle chase:
- $\angle DYC = \angle YDB+\angle DBY = 2\angle DBY = 2 \angle DBC = 2 \angle DAC = \angle DAX+\angle XDA = \angle DXC$.

Now, $\angle DYC = 2 \angle DBC$ is constant beacuse the length of CD is fixed. Then the radius of $XYCD$, which is $\frac{1}{2}\frac{DC}{\sin DYC}$ is constant.

We have $XY = 2R \cdot \sin XCY = 2R \cdot \sin ACB$ = constant, because AB is fixed.

Nice problem, and good night!  :)

Nice ;) in fact now I see that both the problem and your solution are, actually, slightly flawed.

You claim that < DBC is constant because the length of CD is fixed. In fact this is wrong: If R is the radius of the circle S, then $CD=2R\sin\measuredangle DBC$, so that the angle < DBC is either $\arcsin\frac{CD}{2R}$ or $180^{\circ}-\arcsin\frac{CD}{2R}$. Both values are actually possible.

The problem should be fixed as follows:

[color=blue][b]Corrected problem.[/b] Let $A$ and $B$ be two fixed points on a fixed circle $S$, and let $\phi$ be a fixed angle. Two variable points $C$ and $D$ lie on the circle $S$ such that $\measuredangle CAD=\measuredangle CBD=\phi$ and the quadrilateral $ABCD$ is convex. Two points $X$ and $Y$ are chosen on the rays $AC$ and $BC$ such that $AX=AD$ and $BY=BD$. Prove that the distance $XY$ remains constant while the points $C$ and $D$ vary.[/color]

My [i]solution[/i] (somewhat too long, but I wrote it down two years ago when I wasn't particularly experienced in short proofs): Let R be the radius of the circle S. Let M be the midpoint of the arc CD of the circle S (actually, there are two arcs CD, but we take the one which doesn't contain the points A and B). Then, MC = MD. Hence, triangle CMD is isosceles, so that < CDM = < DCM. Since the points C, D, A, M lie on one circle (namely, on the circle S), we have < CAM = < CDM, and similarly < DAM = < DCM. Thus, < CAM = < DAM. Hence, the line AM bisects the angle CAD. Thus, $\measuredangle CAM=\frac{\measuredangle CAD}{2}=\frac{\phi}{2}$. Consequently, by the extended law of sines for triangle CAM (whose circumradius is R), we have $MC=2R\sin\measuredangle CAM=2R\sin\frac{\phi}{2}$.

Since < CAM = < DAM, the line AM is the angle bisector of the angle XAD. But from AX = AD, it follows that triangle XAD is isosceles, and hence, the angle bisector of the angle XAD is the perpendicular bisector of the segment XD. Hence, the line AM is the perpendicular bisector of the segment XD. Thus, MX = MD. Since MC = MD, this becomes MX = MC. Similarly, MY = MC. Hence, MX = MC = MY, and we see that the point M is the circumcenter of triangle CXY. The circumradius of this triangle is therefore $MC=2R\sin\frac{\phi}{2}$.

Hence, after the extended law of sines in this triangle,

$XY=2\left(2R\sin\frac{\phi}{2}\right)\sin\measuredangle XCY=4R\sin\frac{\phi}{2}\sin\measuredangle XCY$.

Now, either < XCY = < ACB or < XCY = 180° - < ACB; in either of these cases, $\sin\measuredangle XCY=\sin\measuredangle ACB$. By the extended law of sines in triangle ABC (whose circumradius is R), we have $AB=2R\sin\measuredangle ACB$. Thus,

$XY=4R\sin\frac{\phi}{2}\sin\measuredangle XCY=4R\sin\frac{\phi}{2}\sin\measuredangle ACB$
$=2\cdot 2R\sin\measuredangle ACB\cdot\sin\frac{\phi}{2}=2\cdot AB\cdot\sin\frac{\phi}{2}$.

Hence, XY is a constant value, qed.

  Darij

Darij, this problem is even more beautiful than you think!  :)

I have just discovered that not only my solution is flawed, but it's the solution of an other problem!

In fact:
- The original problem, solved by you, says that $AX = AD$
- But in my figure, you can see that I took X such that $AX = XD$

Now you may ask: what happens if $AD = XD$?  :D  The answer is: the problem is still true!  :rotfl:

The proofs are not the same, but almost the same.

You can see that $\angle CYD = \angle CXD = f(\angle CAD)$, where f(x) is:
- in the first case, $90+\frac{x}{2}$
- in the second case, $2x$
- in the third case, $x$.

In every problem we have $\angle XCY = \angle ACB$.
Finally, the radius of CDXY is constant, the angle $\angle XCY$ is constant, then $XY$ is constant (in every problem).

Some observations:
- in the first case, triangle $XDY$ is similar to $ADB$
- in the second case, also O passes through the circumcircle of CDXY
- in the third case, the circumcircle of CDXY is the reflection of the circumcircle ABCD wrt. line CD. Then their radii are equal, and XY is constant independently on the length of CD!

bye bye

[quote="edriv"]I have just discovered that not only my solution is flawed, but it's the solution of an other problem![/quote]

Cool :D , actually I should have read your solution more carefully. It is an almost correct solution (the angle < DYC is not fixed, but you only need the absolute value of its sine and this is fixed) for the AX = XD problem.

  darij


I got this question wrong so I did it again.. Is this right?

Problem:

Calculate $\Delta Hr$ when solid copper(I) sulfide reacts with solid sulfur to produce copper(II) sulfide. Write the thermochemical equation for this reaction.

[hide="Solution"]

Balancing equation, I get:

$Cu_2 S + S \rightarrow 2 Cu S$

Then apply Hess's Law

$Cu_2 S \rightarrow 2 Cu + S \ \Delta H = +79.5 \frac{kJ}{mol} \times 1 mol = +79.5 kJ$

$2 Cu + 2 S \rightarrow 2 Cu S \ \Delta H = -53.1 \frac{kJ}{mol} \times 2 mol = -106.2 kJ$

---------------------------------------------------------------------------------------------

$Cu_2 S + S \rightarrow 2 Cu S \ \Delta Hr = -26.7 kJ$

$\therefore$ The thermochemical equation is $\boxed{Cu_2 S + S \rightarrow 2 Cu S + 26.7 kJ}$ [/hide]

Is this right? Hopefully..  :roll:

[quote="wizard2378"]No.

2Cu + S is not equivalent to 2Cu + 2S.

You need to have all the elements and their phases the same to get the heat of reaction correct.[/quote]

Which line are you talking about? I don't see anything wrong with my solution. First line is correct and second line, sulfur is not diatomic so it's just 2S. Huh?  ;)

It looks alright, assuming that the phases all match up (you don't show them).


Prove the identity: (cosx)(tanx+sinxcotx)=sinx+cos^2x

I say there is a typo with the right side of the equation.

MY WORK:

On the left side we have: (cosx)(tanx+sinxcotx)

I know that tanx = sinx/cosx

I also kow that cotx = cosx/sinx

I plugged those "values" into the left side of the equation and simplified.

(cosx)(sinx/cosx) + sinx(cosx/sinx)

cosx cancels out leaving sinx + sinx(cosx/sinx)

Then sinx cancels out leaving sinx + cosx on the left side.

However sinx + cosx does not equal the right side.

What did I do wrong?

You did not multiply the last term in the bracket by $ \cos x$.  :wink:


Given a,b,c are lenght three side of triangle .Prove that:
$ (a \plus{} b \plus{} c)^3  \le \frac{{27a^2 b^2 c^2 }}{{(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)}}$

Let $ a = x + y,b = y + z,c = z + x$. Then our inequality is equivalent to
\[ 8\cdot 8xyz(x + y + z)^3\le 27(x + y)^2(y + z)^2(z + x)^2.
\]
Let $ x + y + z = 1$. Then it is equivalent to
\begin{align*}(x + y)(y + z)(z + x) & \ge\frac {8}{3\sqrt 3}\sqrt {xyz} \\
\Leftrightarrow xy + yz + zx & \ge\frac {8}{3\sqrt 3}\sqrt {xyz} + xyz.\end{align*}
But we have
\[ xy + yz + zx\ge\sqrt {3xyz}\Leftrightarrow \sum_{cyc}x^2(y - z)^2\ge 0.
\]
Let $ t = \sqrt {xyz}$. Then using the above inequality it remains to show that
\[ \sqrt 3t\ge\frac {8}{3\sqrt 3}t + t^2\Leftrightarrow t\le\frac {1}{3\sqrt 3},
\]
which follows from AM-GM:
\[ t^2 = xyz\le\left(\frac {x + y + z}{3}\right)^3 = \frac {1}{27}.
\]

$ (a \plus{} b \plus{} c)^3 \le \frac{{27a^2 b^2 c^2 }}{{(a \plus{} b \minus{} c)(b \plus{} c \minus{} a)(c \plus{} a \minus{} b)}}$
$ \Leftrightarrow 16S^2 (a \plus{} b \plus{} c)^2  \le 27a^2 b^2 c^2$
$ \begin{array}{l}
  \Leftrightarrow (a \plus{} b \plus{} c)^2  \le 27\frac{{a^2 b^2 c^2 }}{{16S^2 }} \\
  \Leftrightarrow (a \plus{} b \plus{} c)^2  \le 27R^2  \\
  \Leftrightarrow \sin A \plus{} \sin B \plus{} \sin C \le \frac{{3\sqrt 3 }}{2} \\
 \end{array}$
 which is obvious
 :)


Does anyone know the names of the national gold medallists who are being trained for the iranian IOI team this year?

do you understand iranian?  if you do, i might be able to help you... otherwise...

Shayan Oveisgharan from Isfahan got gold medal
And the others got silver

That would be last year's result. I am looking for the national gold medallists (in the iranian national olympiad of informatics) this year that are candidates for next year's IOI team.

Hi, I'm an Iranian in Canada and I am wondering how you get into these competitions. Any way I could get in?

In Canada, you need to do the Canadian Computing Competition. It's one of the waterloo contests and your school needs to register for it by January or so. Then, if you do good on that, you advance to the 2nd stage, which is held in Waterloo, and if you get top  4 in that one, you get into IOI and get to go to Poland or some other country where the international competition is held at.

[quote name="bestzack66" url="/community/p17303946"]
[quote=BananaeGod]reported @above[/quote]

wdym
[/quote]

My guess at what happened:

1. Post #7 did not follow the ToS
2. BananaeGod reported it
3. It got deleted

post #7 was a bump
i reported it
it got deleted


A running track is the ring formed by two concentric circles. It is $ 10$ feet wide. The circumference of the two circles differ by about:
$ \textbf{(A)}\ 10\text{ feet} \qquad\textbf{(B)}\ 30\text{ feet} \qquad\textbf{(C)}\ 60\text{ feet} \qquad\textbf{(D)}\ 100\text{ feet} \\
\textbf{(E)}\ \text{none of these}$

The $ 10$-foot size of the track represents the difference in the two concentric circles' radii, meaning that the difference in the diameter would be $ 2*10\equal{}20$ feet. Since circumference is $ \pi d$, the difference in the circumferences would be about $ 60$ feet which is $ \fbox{C}$, because $ \pi$ is around $ 3$, and $ 60$ is a significantly close approximation.

A question, I chose $\boxed{E}$ while taking the test because I got $20\pi$.
Am I not technically right, like what answer gets credit


calculus
integration
inequalities
trigonometry
logarithms
symmetry
real analysis

I'm very sorry for a mistake,everyone.I will correct as follows.

Prove the following inequality.

\[\int^{4007\pi}_{\pi} \frac{|\sin x|}{x} dx>\frac{2}{\pi}\ln2004\]

Write the integral as

$\sum_{k=1}^{n}\int_{k\pi}^{(k+1)\pi}\frac{|\sin x|}xdx = \int_0^{\pi}\left(\sum_{k=1}^n\frac1{x+k\pi}\right)\sin xdx$

where $n=4006$.

We can estimate that for $x<\pi, \sum_{k=0}^n\frac1{x+k\pi}>\frac1{\pi}\sum_{k=1}^n\frac1{k+1}.$

An exercise in comparing boxes to curves leads to the conclusion that

$\sum_{k=1}^n\frac1{k+1}>\int_2^{n+2}\frac1xdx = \ln\left(\frac{n+2}2\right)$

Thus the integral in question is greater than

$\frac1{\pi}\ln\left(\frac{n+2}2\right)\int_0^{\pi}\sin xdx=\frac2{\pi}\ln\left(\frac{n+2}2\right).$

If $n=4006$, we get the integral is greater than $\frac2{\pi}\ln(2004)$

But this estimate isn't very tight. We could also use the symmetry of $\sin x$ in the interval $[0,\pi]$ to claim that

$\int_0^{\pi}\left(\sum_{k=1}^n\frac1{x+k\pi}\right)\sin xdx > \frac2{\pi}\sum_{k=1}^n\frac1{k+\frac12}.$

This is a tighter estimate. However, either estimate is more than strong enough to prove that $\int_0^{\infty}\frac{\sin x}x dx$ does not converge absolutely.

Thank you for your reply, Kent Merryfield.

Exactly, my answer is the same as yours.Especially,I think the latter of your opinion is  the very important.

P.S.  I deleted $n=2,3,\cdot\cdot\cdot$

kunny


If, for RSI, your field is Physics: Astronomy, should you do Physics and Chemistry of Physics and Mathematics?

Thanks,
sran

Since the forms didn't specifically say anything about choosing classes most related to your research, my bet is that you can pick whichever you're more interested in.  There's only a week of classes, right? -- no matter how fast they go, it seems unlikely that you'll learn all the specific chem or math you'll need just during class.

But that should be probably be regarded as my... one cent, it doesn't even qualify as two... since I'm completely unqualified to answer.


function
real analysis
real analysis unsolved

Let $M \subset X$. Prove that:
 If $sup_{x \in M}|f(x)| <+\infty, \forall f \in X^{*}$ then M is bounded.

[quote="Hawk Tiger"]What the meaning of $X^{*}$?? :blush:[/quote]
$X^{*}= L(X,K)$- set of all continnuos linear functions


[quote="kunny"]$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)=1-2(ab+bc+ca)$ so,

$\frac{a^2}{b+ac-b^2}+\frac{b^2}{c+ba-c^2}+\frac{c^2}{a+cb-a^2}$.

$\geq \frac{(a+b+c)^2}{(b+ac-b^2)+(c+ba-c^2)+(a+cb-a^2)}$

$\geq \frac{1}{3(ab+bc+ca)}$

$\geq \frac{1}{(a+b+c)^2}\ [\because (a+b+c)^2\geq 3(ab+bc+ca)\ ]$[/quote]

I think u used this lemma: if $a,b,c$ are reals and $x,y,z>0$ then we have:
$\sum \frac{a^2}{x} \geq \frac{(a+b+c)^2}{x+y+z}$. But this is true iff $x,y,z>0$. So, u have to prove
that $b+ac-b^2 > 0$,$c+ba-c^2 > 0$,$a+cb-a^2 > 0$ to be your sollution true.    :roll:

[quote="paladin8"]Well since $a < 1$, $a-a^2 > 0$ and same for $b,c$ so it's true.[/quote]

right  ;)

[quote="jianuovidiu"]Prove that for any positive real numbers $a,b,c$ with $a+b+c=1$ the following inequality is true: $\frac{a^2}{b+ac-b^2}+\frac{b^2}{c+ba-c^2}+\frac{c^2}{a+cb-a^2}\geq 1$.[/quote]
$b+ac-b^2=ab+bc+ca$
$a^2+b^2+c^2 \ge ab+bc+ca$ is simple.


[quote="khashi70"]$ p \equal{} a^{2} \plus{} 14b^{2}$ is a prime number . prove that the following equation has an integer solution :
$ (a^2 \plus{} 1)^{2} \equiv 8 (mod p)$[/quote]
you want write the equation $ (x^2\plus{}1)^2\equiv 8(mod\ p)$


Let [tex]x_1,x_2,x_3,x_4[/tex] are real numbers such that
[tex]x_1^2+x_2^2+x_3^2+x_4^2=1[/tex]
Find greast valued and least valued of
[tex]x_1^2+x_2^2+2x_1x_3+3x_2x_3+x_3x_4[/tex]

[quote="harazi"]I think, for this kind of problems a matricial approach is best.[/quote]
What is your solution?


The set of all positive integers is the union of disjoint subsets {f(1), f(2), . . .} and
{g(1), g(2), . . .} where
f(1) < f(2) < . . . ,
g(1) < g(2) < . . . ,
g(n) = f(f(n)) + 1 for all n>= 1. Determine f(2007)

This is problem is very closely related to Wythoff's Nim.  Here is a link:

http://www.google.com/search?hl=en&q=wythoff%27s+nim

Additionally, if $ \phi$ is the positive real such that $ \phi^{2}\equal{}\phi\plus{}1$, then the explicit formulae for these functions are $ f(n) \equal{}\lfloor\phi n\rfloor$ and $ g(n)\equal{}\lfloor\phi^{2}n\rfloor$, thus making $ f(2007)\equal{}3247$.  These are Beatty sequences, so it is obvious they partition the positive integers, and the references above about the context of the game show that the relation between $ f$ and $ g$ holds.  Here is another couple sites that expound on this interesting topic:

http://www.cut-the-knot.org/arithmetic/whole_part.shtml
http://en.wikipedia.org/wiki/Beatty_sequence


abstract algebra
superior algebra
superior algebra solved

Hi,

I have tried to prove this problem with no luck.  I would like to see how it is proven.

Given a Artinian right R-module Q prove that each homomorphic image of Q is right Artinian.

Thank you,
Sussi

Hi Sussi,

it's quite obviuos: Let $f : Q \to P$ be an epimorphism, and

$P_{0}\supseteq P_{1}\supseteq P_{2}\supseteq ...$

an descending chain of submodules of $P$. Now consider the submodules $Q_{i}: = f^{-1}(P_{i})$ of $Q$, and use that $Q$ is Artinian (and use the surjectivity of $f$ to show $P_{i}= f(Q_{i})$).

-oo-

-oo-,

I know that I must be thick but can I just state that it is surjective (and thus an epimorphism)?

Since Q is Artinian there is a descending chain of submodules that becomes constant.  Since Q_i = f^-1 ( P_i), this also holds for  f^-1 ( P_i).  Would not f(Q_i) = P_i automatically since it is surjective or am I incorrect?  And it follows from this?

I am sorry.  I am having a lot of difficult with modules.

Sussi

You reverse the implications :/

f is an epimorphism; [i]thus[/i] f is surjective.

Since $Q$ is artinian, [i]every[/i] descending chain becomes constant. This holds especially for the stated submodules $Q_{i}$.

Indeed, $f(Q_{i}) = P_{i}$ is a consequence of the fact, that $f$ is surjective. I've alread mentioned this above.

Ok, so am I right now or still misunderstanding?

Proof:   Let f:  Q -> P be an epimorphism  (I messed up my question -- "Can I state this is an epimorphism outright"?).  Let P0 containing P1 containing P2 .... be an descending chain of submodules of P.  THe submodules Q_i = f^-1(P_i).  We are given that Q is Artinian, so for some "r" Q_r becomes constant and thus f^'1(P_r) is constant as well due to equality.  Since f is onto, P_i = f((Q_i)) and so P_r = f((Q_r)) and P_r is constant as well.  So every homomorphic image of Q is Artinian.

Sussi


Nice solution Virgil Nicula, that's what I did.

Minsoens, you had a good solution too :D


Conculate sum:[tex]f(x)=\sum_{n=1}^ {\infty}F_n.x^n[/tex],
with [tex]F_n[/tex] is n-th Fibonaxi number

Did I miss something or it is trivially $\frac x {1-x-x^2}$ ?  ;)

Pierre.


The problems is:
[quote]
Find the volume and surface area of a hemisphere with a radius of 8cm when a sector of 60 degrees has been removed.
[/quote]

and I was able to get the surface area, but how do I find the volume of the sector? would it be like finding the area of it?

I wouldn't even find the volume of the sector, because now all you have to do is find the volume of 5/6 of the hemisphere.

so then how would I find the volume of the hemisphere with a 60 degree sector missing?

[quote="fishythefish"]Read what I just said.

Because 60 is one-sixth of 360, the sector is one-sixth the volume of the hemisphere. When you subtract, you find that all you need to do is find five-sixths of the volume of the hemisphere.[/quote]Not quite correct... that would be for a circular sector, not a spherical sector :ninja:

What exactly is a spherical sector anyways? And what would a 60-degree one look like?

Okay, but it doesn't make a difference...I think.... :maybe:

This is the way I see it (correct me if I'm wrong).
Imagine you are looking from a bird's-eye view over the center of the hemisphere. If a 60 degree sector is taken out, wouldn't that just leave you with 5/6ths of the hemisphere :?:


1. In "My USAMTS", I click "My Profile".
2. Then I go to the dropdown where it says Grade and it's at "None Selected". I click on it and pick "10".
3. Then I scroll down and click submit changes.

4. Then it shows a page with the "10" changed back to "None Selected". However, if I made any other changes, it saves it.
5. I go back to step 2, and step 4 shows the same thing.

What's wrong?

I don't know...I can't reproduce the error.

What browser type and version are you using?

[quote="ProtestanT"]I'm using Firefox 2.0.0.16.

Also, I just tested it with IE 6, and it doesn't work either.[/quote]Those are both rather old browsers.  Try IE 7 or 8 or Firefox 3.

I'll have our programmer look into it on Monday.

I just upgraded my browser to Firefox 3. The same problem is still there.

Does there need to be something filled for "School CEEB code" or "Teacher Name"? I have those blank.

Well I don't have school code or teacher name in....and I can still see the grade.

I'm betting it has something to do with your cookie settings...


I really want to prepare for the AMC 12 this coming year, but I have a few questions.
Are the Mock AMC's (all 10 of them or so) accurate? Should I just use them to practice, or should I purchase the ones from UNL. AMC is coming up pretty soon, so the tests might not even have time to come. So basically, are the Mock AMC's good enough to practice with?

Also, are the Mock AMC's 10 or 12?

Thanks in advance! :)

Not sure what do you mean by Mock AMC.

Do you mean past years' mock AMC?

Then it's yes.

Does it have enough time to come?

If you take B, yes. If you take A, possibly but I can't guarantee for sure.

Most of them are 12 with the occasional "11" and maybe one "10" from what I've gathered.

They are a bit harder than what you can expect to see but hey... that's good for practice isn't it? Challenge is important.

I would do them if only for the sole reason that you will score slightly lower on the Mock ones then on the real ones and then you will be happier with whatever score you will actually get since it will be higher than your Mock scores.

It's really hard to answer the questions you are asking.  First of all, enough for what?  What are you trying to accomplish by practicing?  You have to be more specific.  You could be trying to make AIME, trying get a decent score for a good USAMO index,  trying to ace it (i.e. 150), or just trying to get a personal best.  Each of these require different levels of preparation not to mention that each person learns differently and would thus have to train differently.  It also highly depends on your current ability and how easily you learn.

There is not much more you can do between now and February that would dramatically change your score.  But what I would do is focus on your weakest area and take a few practice tests to see how you're doing.

Good luck!

[quote="vinny919"]So, considering that there I will get very little practice time with the real tests, should I even order them??

Is practicing with the Mock AMC's (that are listed on a thread at the top of this board) sufficient?[/quote]

You can find old tests online

[url]http://www.math.uncc.edu/~hbreiter[/url]

[url]http://www.math.ksu.edu/main/events/hscomp/samples/amc12/sample.htm[/url]

But the Mock AMCs are good practices too!


Let $ a,b,c,d\in[\minus{}1,1]$ such that $ a\plus{}b\plus{}c\plus{}d\equal{}0$. Find the minimum and maximum values of $ f(a,b,c,d)\equal{}\sqrt{1\plus{}a}\plus{}\sqrt{1\plus{}b}\plus{}\sqrt{1\plus{}c}\plus{}\sqrt{1\plus{}d}$.

$ max \equal{} 4$ obviously.
$ min$ can be found by Lagrange's method( :D )...


trigonometry
parameterization
integration
function
calculus
calculus computations

Find the length of the parametric curves x = cos(e^t), y = sin(e^t) for 0 <=t<=1. Explain why your answer is reasonable.

I got the answer as e-1, but can't answer why this is reasonable. Can someone help me??

I assume that's supposed to be $x=\cos(e^{t}).$

This is simply a non-standard parameterization of an arc of the unit circle. Make the substitution $\theta=e^{t}$ and you'll see that $\theta$ goes from $1$ to $e.$ So that change of angle is (by the principle of radian measure) the arclength.

I know how to get to e-1 but I mean by just looking at the function given first time without having to calculate it first, how do I argue that this is possible

EquinoX, did you not read my post above? Take your fingers off the keyboard and spend some time thinking.

I read your post but I didn't get it, can you explain it easier

Do you know what it means to define a curve parametrically?  And, if so, can you explain exactly which parts of what Kent wrote you didn't understand?

I do understand on what is it to define a curve parametrically. If I were drawn a picture of this parameter equation then this would be represented as the circumference of 1/4 of a circle right?? I dont understand if I make substitution of theta of e^t then how will I see that theta goes to 1 to e?

It would be some portion of a circle, yes.  It's not clear to me why you would think that it would be one quarter.

We agree that the equations $x = \cos \theta$ and $y = \sin \theta$ parametrically define some portion of a circle.  The collection of values that $\theta$ takes on determines exactly what portion of that circle.  In our case, we have something more complicated, because $\theta$ is determined as a function of $t$.  So, in order to figure out the range of values that $\theta$ takes, you need to look at the expression for $\theta$ in terms of $t$ and the range of values that $t$ takes.

As a general rule: it's much easier to help you when you make it very clear what your problem is.  Detailed questions or explanations of difficulty, as in your most recent post, are much easier to answer in a satisfactory manner than vage, general questions ("can you explain it easier?").


quadratics
modular arithmetic
number theory proposed
number theory

Prove that if $\{F_n\}$ denotes the sequence of the Fibonacci numbers and if $p$ is a prime then $p$ divides \[F_p - \left(\frac{p}{5}\right).\]
[I tried to create a Legendre symbol here, doesn't look exactly as it should look but anyway...]

While I think the problem should not be difficult once it's stated properly, you really need to say what your $F_0,F_1$ are.

[quote="nhat"]oh it's very easy [...] and newton's lemma.  :D  :D  :D[/quote]
Really? Can you give me some details??? :huh:

I was thinking about a similar approach, but someone really needs to tell me what the first values of $(F_n)_n$ are in this case... :)

For the classical Fibonacci sequence, we have: $F_1=F_2=1$. No?
Could somebody explain to me what is [b]Newton's lemma[/b]? Thanks!

Oh, I was making a mistake (not the first time I do it, you know :)), so I thought $F_1=1,F_2=1$ didn't work :).

I don't know what Newton's Lemma is, but here's what I did:

Let $u=\frac{1+\sqrt 5}2,\ \bar u=\frac{1-\sqrt 5}2$. We have $F_p=\frac 1{\sqrt 5}\cdot (u^p-\bar u^p)\ (*)$. If $5$ is a quadratic residue $\pmod p$, then we can replace $\sqrt 5$ with an actual residue $\pmod p$, so we can write $(*)$ as $F_p=1\pmod p$, which is precisely what we want.

On the other hand, if $5$ is not a quadratic residue $\pmod p$, we can interpret $(*)$ as a relation in $\mathbb Z_p[\sqrt 5]$, which is a field with $p^2$ elements, in this case. Rewrite $(*)$ (we're working in the field $\mathbb Z[\sqrt 5]$) as $F_p=\frac 1{\sqrt 5}\cdot (u-\bar u)^p=\sqrt 5^{p-1}=5^{\frac{p-1}2}=-1$, which is the desired result in this case.

You are of genius, as always. :)
Unfortunately, I didn't understand the second part. :blush:  :blush:  :blush:

Are you referring to the part where $\left(\frac 5p\right)=-1$? (actually, it's $\left(\frac p5\right)=-1$, but it's the same thing by Quadratic Reciprocity)

Since $5$ is not a quadratic residue, we can adjoin $\sqrt 5$ to the field $\mathbb Z_p$ and we obtain a field with $p^2$ elements, namely $\mathbb Z_p[\sqrt 5]$. In any field with characteristic $p$ (and $\mathbb Z_p[\sqrt 5]$ is such a field), we have $a^p+b^p=(a+b)^p$. We apply this to $a=u,b=-\bar u$ to get $u^p-\bar u^p=(u-\bar u)^p$. $u-\bar u=\sqrt 5$, so $F_p=\frac 1{\sqrt 5}\cdot (u-\bar u)^p=\frac 1{\sqrt 5}\cdot \sqrt 5^p=\sqrt 5^{p-1}$. This is $5^{\frac {p-1}2}$, which is $-1\pmod p$ because $5$ is not a quadratic residue.

I hope it's clearer now (actually, to say the truth, I hope it's correct, and, if it's correct, I hope it's clear :)).

[quote="grobber"]Since $5$ is not a quadratic residue, we can adjoin $\sqrt 5$ to the field $\mathbb Z_p$ and we obtain a field with $p^2$ elements, namely $\mathbb Z_p[\sqrt 5]$.[/quote]
I don't know this theory.

[quote="grobber"]we're working in the field $\mathbb Z[\sqrt 5]$[/quote]
For me, $\mathbb Z[\sqrt 5]$ isn't a field (only a ring) :?

I think he said $\mathbb{Z}_p [\sqrt5] =\mathbb{Z} [\sqrt5] / p$ and this is a field

Yes, by $\mathbb Z_p$ I mean the field with $p$ elements. It's not hard to show that if $5$ is not a quadratic residue of $p$, then $\mathbb Z_p[\sqrt 5]=\mathbb Z_p(\sqrt 5)$, meaning that the ring formed by adjoining $\sqrt 5$ to $\mathbb Z_p$ is, in fact, a field with $p^2$ elements.

We can use a more elementary redaction (We don't need splitting field) :

[color=red][[b]Moderator edit:[/b] In the following solution, $\binom{k}{p}$ should be read as $\binom{p}{k}$].[/color]

Let $p>2$ a prime, $\alpha=\frac{1+\sqrt{5}}{2}$ and $\beta=\frac{1-\sqrt{5}}{2}$.
$5^{\frac{p-1}{2}}=\frac{\sqrt{5}^p}{\sqrt{5}}=\frac{(\alpha-\beta)^p}{\sqrt{5}}=F_p+p\sum^{p-1}_{k=1}\frac{1}{p}\binom{k}{p}\alpha^k(-\beta)^{p-k}$
If $0<k<p$, we have $p|\binom{k}{p}$, so:
$F_p-5^{\frac{p-1}{2}}=\frac{p}{2^p}(a+b\sqrt{5})$, with $(a,b)\in\mathbb{Z}^2$
Thus $b=0$ and $ap=2^p(F_p-5^{\frac{p-1}{2}})\Rightarrow p|F_p-5^{\frac{p-1}{2}}$.