Difference between revisions of "2021 IMO"

(Created page with "합 q+r= (x^2+y^2-z^2)/2")
 
Line 1: Line 1:
q+r=  (x^2+y^2-z^2)/2
+
For the statement to be true, there must be at least three pairs whose sum is each a perfect square.
 +
There must be p,q,r such that p+q = x^2 and q+r = y^2 p+r = z^2.
 +
WLOG n<= p<= q<= r <= 2n ... Equation 1
 +
p = x^2 + z^2 - y^2
 +
q = x^2 + y^2 – z^2
 +
r = y^2 + z^2 – x^2
 +
by equation 1
 +
2n <= x^2 + z^2 – y^2 <= 4n
 +
2n <= x^2 + y^2 – z^2 <= 4n
 +
2n <= y^2 + z^2 – z^2 <= 4n
 +
 
 +
6n <= x^2 + y^2 + z^2 <= 12n
 +
  6n <= 3x^2 <= 12n
 +
2n <= x^2 <= 4n
 +
{\displaystyle {\sqrt {\quad }}}2n <= x <= 2 * {\displaystyle {\sqrt {\quad }}}n
 +
At this time n >= 100, so
 +
 
 +
10 * {\displaystyle {\sqrt {\quad }}}2 <= x,y,z <= 20
 +
15 <= x,y,z <= 20
 +
where 2|x^2 + y^2 – z^2 , 2|x^2 + z^2 – y^2 , 2|y^2 + z^2 – z^2
 +
x = 16, y = 18, z = 20 fits perfectly
 +
at least the proposition is true

Revision as of 10:59, 29 January 2023

For the statement to be true, there must be at least three pairs whose sum is each a perfect square. There must be p,q,r such that p+q = x^2 and q+r = y^2 p+r = z^2. WLOG n<= p<= q<= r <= 2n ... Equation 1 p = x^2 + z^2 - y^2 q = x^2 + y^2 – z^2 r = y^2 + z^2 – x^2 by equation 1 2n <= x^2 + z^2 – y^2 <= 4n 2n <= x^2 + y^2 – z^2 <= 4n 2n <= y^2 + z^2 – z^2 <= 4n

6n <= x^2 + y^2 + z^2 <= 12n

6n <= 3x^2 <= 12n

2n <= x^2 <= 4n {\displaystyle {\sqrt {\quad }}}2n <= x <= 2 * {\displaystyle {\sqrt {\quad }}}n At this time n >= 100, so

10 * {\displaystyle {\sqrt {\quad }}}2 <= x,y,z <= 20 15 <= x,y,z <= 20 where 2|x^2 + y^2 – z^2 , 2|x^2 + z^2 – y^2 , 2|y^2 + z^2 – z^2

x = 16, y = 18, z = 20 fits perfectly
at least the proposition is true