Difference between revisions of "2021 IMO"

Line 5: Line 5:
  
 
p = x^2 + z^2 - y^2
 
p = x^2 + z^2 - y^2
 +
 
q = x^2 + y^2 – z^2
 
q = x^2 + y^2 – z^2
 +
 
r = y^2 + z^2 – x^2
 
r = y^2 + z^2 – x^2
  
Line 11: Line 13:
  
 
2n <= x^2 + z^2 – y^2 <= 4n
 
2n <= x^2 + z^2 – y^2 <= 4n
 +
 
2n <= x^2 + y^2 – z^2 <= 4n
 
2n <= x^2 + y^2 – z^2 <= 4n
 +
 
2n <= y^2 + z^2 – z^2 <= 4n
 
2n <= y^2 + z^2 – z^2 <= 4n
 +
  
 
6n <= x^2 + y^2 + z^2 <= 12n
 
6n <= x^2 + y^2 + z^2 <= 12n
 +
 
  6n <= 3x^2 <= 12n
 
  6n <= 3x^2 <= 12n
 +
 
2n <= x^2 <= 4n
 
2n <= x^2 <= 4n
 +
 
√(2n) <= x <= 2√n
 
√(2n) <= x <= 2√n
 +
 
At this time n >= 100, so
 
At this time n >= 100, so
  
 
10 * √2 <= x,y,z <= 20
 
10 * √2 <= x,y,z <= 20
 +
 
15 <= 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
+
 
 +
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
 
  x = 16, y = 18, z = 20 fits perfectly
  at least the proposition is true
+
 
 +
  therefore the proposition is true

Revision as of 10:02, 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

√(2n) <= x <= 2√n

At this time n >= 100, so

10 * √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
therefore the proposition is true