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#51 Re: Help Me ! » A proof I'm looking for » 2009-04-25 00:13:00
maybe this list can help: http://agutie.homestead.com/files/geometry_help_online.htm
edit: maybe cevas theorem of menelaos theorem?
http://en.wikipedia.org/wiki/Ceva%27s_theorem
http://en.wikipedia.org/wiki/Menelaus%27_theorem
#52 Exercises » Kurre's Exercises » 2009-04-17 23:33:01
- Kurre
- Replies: 33
#1 Solve:
a)
b)
#2 Find all functions satisfying:
#3 Find all functions satisfying:
#4 Find all functions satisfying:
#5 Let a be a given real number. Find all functions satisfying:
#6 Find all injective functions satisfying:
#53 Re: Help Me ! » groups » 2009-04-08 04:26:30
Im not sure what conjugate means, but I can maybe guess it means that x^2=1. Because that can be proved using Lagrange theorem.
#54 Re: Help Me ! » rearranging to find IRR » 2009-04-05 21:12:24
I think it is much easier to solve the quadratic for t=1/(1+IRR) directly, then after that solve for IRR.
#55 Re: Exercises » Guided exercises » 2009-03-31 20:50:00
OR you can use conjugate rule
#58 This is Cool » Geometry website » 2009-03-22 22:06:09
- Kurre
- Replies: 1
Wonderful page if you want to practice geometry:
http://agutie.homestead.com/files/index.html
includes tons av theorems and problems, animated proofs, puzzles, history and a lot more.
Enjoy!
#59 Re: Exercises » three distinct » 2009-03-22 04:55:39
Now we multiply (1) by b, (2) by c and subtract, and also we subtract (3) multiplied by a from (1) multiplied by b, yielding:
Now we also multiply (2) by ac, (1) by c b and subtract, yielding:
inserting (a-c) in (4) from (6) yields:
and multiplying this by a/t (t nonzero) and using (5)
since the numbers where distinct, , thus
#60 Puzzles and Games » functional equation » 2009-03-22 02:54:16
- Kurre
- Replies: 1
Solve the functional equation f(az)=af(z), where a is some complex constant with |a|>1, and f(z) is continuous in a domain D.
I have managed to solve the problem where f is analytic at the origin, but I cant manage to solve it when f is only continuous.:/
#61 Re: Help Me ! » matrices » 2009-03-21 23:10:37
This problem has been annoying me for a while, and thats not weird since its not true! 
assume that A= D, C=B=I are diagonal matrices. Then AB^t and CD^t are diagonal and therefore also symmetric. also let
The each element in AD^t on the diagonal (its a diagonal matrix) is equal to 2, and BC^t=I. So the condition AD^t-BC^t=I holds. But A^tD+C^tB is a matrix with diagonal entries equal to 3, so the theorem is false. bad bad tony!
#62 Re: Puzzles and Games » Find » 2009-03-19 05:54:40
multiplying the two equations yields:
(3)
Now expanding (1) and (2) yields:
.
adding these two equations and using (3) yields:
but √5/2+√2 is greater than 2 which would be impossible (y <4), thus
#63 Re: Puzzles and Games » imo » 2009-03-17 00:05:08
hm I missed this one. I also solved it, dont think it was so difficult. By the way, why the title? is this really an IMO problem? They usually only have discrete mathematics.
#64 Re: Puzzles and Games » Numbers games » 2009-03-12 08:01:19
my first thought on the first one was 75*10/(8-6)+3
#65 Re: Puzzles and Games » Type the username above you - WITH YOUR ELBOW! » 2009-03-11 08:21:13
wes3bjhyuiuter weol,frgt - white wolf 
#67 Re: This is Cool » Simple game » 2009-03-11 08:06:51
What about this
#68 Re: Help Me ! » Power Series Proof » 2009-03-11 07:59:41
Hint:
Go backwards and use that:
#69 Re: Help Me ! » trig simplification » 2009-03-07 21:34:42
Now use the conjugate rule and half angle formula for cosine
#70 Re: Puzzles and Games » m(AED )= 90° » 2009-03-03 02:59:46
I got it. 
Let the points be represented by numbers in a complex plane (a,b,c,d,x,y and s instead of E), and insert the coordinate axes such that D=0, the circumcircle of CXY is the unit circle and BC is parallell with the real axis. Let
|AB|=|AC|=R. The fact that XY||AB implies that XYC also is an isosceles triangle with |XY|=|CY|. From that we can easily conclude that y=i. We also have that:
so
we have that, since BC is parallell to the real axis:
Also, we have that
which yields:
We also have (for the complex number s representing the point E):
Now consider a-s, which represents the vector EA starting from the origin. If we rotate it pi/2 radians counter clockwise, and if AED=pi/2, the two vectors should be paralell. Rotation by pi/2 is just multiplication by i:
Now if this was parallell with DE, there should be a real (positive) constant k such that v=ks. ie Re(v)=kRe(s), Im(v)=kIm(s). so starting with the imaginary parts:
Note also that k is positive. Now we just need to check that the equation for the real parts hold for this k. Indeed:
Thus
Q.E.D
#71 Re: Puzzles and Games » Neat problem » 2009-03-02 06:16:19
Hopefully you can see why using LaTeX makes things far more readable than the pathetic shit written by bitus. I hope bitus goes to hell when he dies (which I also hope will be sooner rather than later)!
hahaha were you drunk when you wrote that??
any possible sarcasm is definitely not clear
#72 Re: Help Me ! » URGENT_Mathematical Algorithms_Please help me » 2009-03-01 07:28:24
6. Isnt the order just with increasing serice time? ie if a is served before b, Ta≤Tb.
#73 Re: Help Me ! » Proof of the sum angle identity » 2009-03-01 06:49:12
A nice geometric construction I found:
http://www.geocities.com/kurre999/Trigaddition.bmp
not a complete proof though since it doesnt work for all angles, but atleast a good explanation for it
edit: actually I think its easy to prove the formula completely using this. the picture works for angles pi/2<x+y<pi. If x+y<pi/2 the geometric construction just changes a bit. We can easily(?) derive the formula for cos(x+y) with the same set of angles x,y. Now if x+y>pi (or x+y<0) we just scale off all multiples of 2pi until we are in 0<x+y<2pi. If we are in 0<x+y<pi we are done, otherwise we use sin(a+pi)=cos(a) and use the addition formula for cos.
edit2: but I realize we get problems since we can only scale of x and y seperately, so it doesnt really work...
edit3: But, using sin(x+y)=-sin(-x-y)=-sin((pi-x)+(pi-y)) if x+y>pi but 0<x,y<pi will do the job
#74 Re: Help Me ! » List of quadrilateral vector proofs » 2009-02-27 23:08:33
One cool theorem, altough maybe not so useful, is ptolemys theorem.
A quadrilateral ABCD is cyclic iff:
|AC||BD|=|AB||CD|+|BC||AD|
http://en.wikipedia.org/wiki/Ptolemy%27s_theorem
#75 Re: This is Cool » Why Z^2=x^2 + Y^2 » 2009-02-27 08:45:28
I guess you are thinking of one of the geometric constructions on this website.
http://jwilson.coe.uga.edu/EMT668/emt668.student.folders/HeadAngela/essay1/Pythagorean.html