Difference between revisions of "2021 IMO"
Mathhyhyhy (talk | contribs) |
Mathhyhyhy (talk | contribs) |
||
Line 1: | Line 1: | ||
For the statement to be true, there must be at least three pairs whose sum is each a perfect square. | 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. | + | 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 | WLOG n<= p<= q<= r <= 2n ... Equation 1 | ||
+ | |||
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 | ||
+ | |||
by equation 1 | by equation 1 | ||
+ | |||
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 | ||
Line 13: | Line 17: | ||
6n <= 3x^2 <= 12n | 6n <= 3x^2 <= 12n | ||
2n <= x^2 <= 4n | 2n <= x^2 <= 4n | ||
− | + | √(2n) <= x <= 2√n | |
At this time n >= 100, so | At this time n >= 100, so | ||
− | 10 * | + | 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 | at least the proposition is true |
Revision as of 10:01, 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 at least the proposition is true