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AsianSensation




PostPosted: Mon Jul 07, 2003 1:36 pm   Post subject: (No subject)

crap, I forgot to give the bits to bugz....oh well, here is the 60 bits that i promised, and your counter starts up new

I haven't been working on that problem, too much critical reading....

anyways, I'll get started soon(but it's not other people are trying for these questions anyways Confused )

+60 bits to Bugz
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AsianSensation




PostPosted: Fri Jul 18, 2003 4:48 pm   Post subject: (No subject)

um........how do you tell if a number is prime?

I know one method, and I've been using it, but it's not giving me results....

Im using Wilson's Thereom, which state that the positive integer n is prime if and only if (n-1)! = -1 (mod n)

otherwise, how else could you check to see if it's prime or not? I have to divide all the previous integer before n/2?
SilverSprite




PostPosted: Fri Jul 18, 2003 4:58 pm   Post subject: (No subject)

.........Wilsons theorem?? Thats actually a theorem? isnt that a bit obvious?
SilverSprite




PostPosted: Fri Jul 18, 2003 4:58 pm   Post subject: (No subject)

And there is no easy way...why do you think ppl use encryptions with prime numbers... they arent easy to find.
AsianSensation




PostPosted: Fri Jul 18, 2003 6:00 pm   Post subject: (No subject)

i was surprised too that it was a thereom, but hey, if it says so it's a thereom, then it is.

so i have to prove all cases involving p^3 + 2 divide by all integers before (p^3 + 2)/2? And show that they cannot be a integer?

btw, if p is prime, and p + 2 is prime, what is that called? twin primes?
bugzpodder




PostPosted: Sat Jul 19, 2003 10:19 am   Post subject: (No subject)

Quote:

so i have to prove all cases involving p^3 + 2 divide by all integers before (p^3 + 2)/2? And show that they cannot be a integer?


no actually, if p is 1 or 2 mod 3, then p^2+2 = 1+2 (mod 3) or p^2+2=4+2 (mod 3) which is always divisble by 3 (for all p>1). so p^2+2 is prime iff p=3, and we can easily verify that 3^3+2=29 is prime
AsianSensation




PostPosted: Sat Jul 19, 2003 12:23 pm   Post subject: (No subject)

ah.....nice, so p=3 is the only case?

anyways...since bugz posted up the solutions, and no one got it, then bugz keep his bits.

I'll load up another question in a minute.

I'll load up a geometry one, it will be the first one I do, so bare with the picture if it's not that good.
AsianSensation




PostPosted: Sat Jul 19, 2003 12:29 pm   Post subject: (No subject)

Question #5:

Let P be the center of the square constructed on the hypotenuse AC of the right-angled triangle ABC. Prove that BP bisects angle ABC.


Edit:
And since I'm working on a different computer that doesn't have any image editing things other than paint, I'll give you the code in turing, just run it and it should show you the picture.

code:

var fontID := Font.New ("Ariel:12:Bold")
drawbox (50, 50, 150, 150, 7)
drawline (50, 150, 15, 70, 7)
drawline (50, 50, 15, 70, 7)
drawline (100, 100, 15, 70, 7)
Font.Draw ("A", 45, 160, fontID, 7)
Font.Draw ("B", 10, 50, fontID, 7)
Font.Draw ("C", 45, 35, fontID, 7)
Font.Draw ("P", 100, 80, fontID, 7)
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Catalyst




PostPosted: Sat Jul 19, 2003 1:17 pm   Post subject: (No subject)

doesnt sound like C should be where it is
bugzpodder




PostPosted: Sat Jul 19, 2003 2:19 pm   Post subject: (No subject)

obviously ABCP is a cyclic quadrilateral (opposite angles add up to 180) and furthermore since ABP=ACP=45 (common chord in the circle). since ABC is 90 then PBC is also 45 and hence BP is the angle bisector
SilverSprite




PostPosted: Sat Jul 19, 2003 6:04 pm   Post subject: (No subject)

AsianSensation wrote:
Let P be the center of the square constructed on the hypotenuse AC of the right-angled triangle ABC. Prove that BP bisects angle ABC.


Bugz likes to shorten his solutions as much as he can. Anyways this is an add-on to the beginning of bugz's solution since he seems to have flew by it. Or maybe I am wasting my time because this is common knowledge and is unnecessary to state but here it is anyways:

Since P is the centre of the square, PA=PC, and angleAPC=90degrees. Hence P lies on the circle. And this is where bugz comes in.
AsianSensation




PostPosted: Sun Jul 20, 2003 11:21 am   Post subject: (No subject)

Yeah, the labeling was wrong, I fixed that.

anyways, here is my solution, would have posted it up yesterday, if I didn't go to that party.

Anyways, Since ABCP is a cyclic quadrilateral, draw a circle that circumscribe ABCP. Since angle PAC = angle PCA = 45, therefore, arc AP = arc PC, therefore, the angle subtended by AP and PC are equal, therefore, angle ABP = angle PBC = 45, therefore, PB bisect angle ABC.

btw, I added an attachment of the picture, it works for both bugz and mine solution.



geometry.jpg
 Description:
 Filesize:  5.83 KB
 Viewed:  2542 Time(s)

geometry.jpg


AsianSensation




PostPosted: Sun Jul 20, 2003 11:23 am   Post subject: (No subject)

ok, for those solutions, since it wasn't a hard problem, I give bugz 10 bits for solution 1, and SilverSprite 5 bits for clarification, additionally, for solution to problem 4, I give bugz 30 bits, check scoreboard for your total amount.
bugzpodder




PostPosted: Sun Jul 20, 2003 12:54 pm   Post subject: (No subject)

short little problem: Prove gcd(nCi,nCj)>1, given 0<i,j<n
bugzpodder




PostPosted: Sun Jul 20, 2003 12:55 pm   Post subject: (No subject)

nCi is obviously n Choose i
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