Author |
Topic: Couple of 2003 Questions (Read 647 times) |
|
ThudnBlunder
Uberpuzzler
The dewdrop slides into the shining Sea
Gender: 
Posts: 4489
|
 |
Couple of 2003 Questions
« on: Dec 7th, 2003, 4:56am » |
 Quote  Modify
|
1) Do there exist polynomials a(x), b(x), c(y), and d(y) such that
1 + xy + x2y2 [smiley=equiv.gif] a(x)c(y) + b(x)d(y)?
2) Let A, B, C be equidistant points on the circumference of a circle of unit radius centred at O, and let P be any point in the circle's interior. Let a, b, c be the distances from P to A, B, C, respectively. Show that there is a triangle with side lengths a, b, c and that the area of this triangle depends only on the distance from P to O.
|
| « Last Edit: Dec 8th, 2003, 7:13am by ThudnBlunder » |
 IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
Eigenray
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 1948
|
|
Re: Couple of 2003 Questions
« Reply #1 on: Dec 7th, 2003, 10:58pm » |
Quote Modify
|
This years problems have already been posted to sci.math. Extremely simple solutions to all the problems, which make you feel like a moron for not getting them, will also appear shortly. I always try to avoid reading them until I can solve them myself.
Damn it, I thought B1 was supposed to be easy, to make you feel better after doing so horribly on part A. I still haven't gotten this one.
For B5, I let the points on the circle be 1, w, w2=w', where w is a primitive cube root of 1 (z' = z-bar), then plugged a=|p-1|, b=|p-w|, c=|p-w'| into Hero's formula, K=sqrt[s(s-a)(s-b)(s-c)], which expands to
16K2 = 2(a2b2+a2c2+b2c2)-(a4+b4+c4 ),
and then did about half a page of algebra to get K = (1-|p|2)sqrt(3)/4. It's not too bad if you make some clever substitutions and keep using that w+w'=-1, ww' = 1.
A1 was easy, A2 was neat. A3 is on this site somewhere, and now I see an infinitely simpler proof than mine already on sci.math. B2 and B3 aren't very hard, and I haven't really thought much about the others yet.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print