Difference between revisions of "2021 IMO"
Mathhyhyhy (talk | contribs) (Created page with "합 q+r= (x^2+y^2-z^2)/2") |
Mathhyhyhy (talk | contribs) |
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− | + | 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 09:59, 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