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Difference between revisions of "2014 USAMO Problems/Problem 3"

(Created page with "==Problem== Prove that there exists an infinite set of points <cmath>\ldots,\,\,\,\,P_{-3},\,\,\,\,P_{-2},\,\,\,\,P_{-1},\,\,\,\,P_0,\,\,\,\,P_1,\,\,\,\,P_2,\,\,\,\,P_3,\,\,\,\,\...")
 
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==Solution==
 
==Solution==
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Consider an elliptic curve with a generator <math>g</math>, such that <math>g</math> is not a root of <math>0</math>. By repeatedly adding <math>g</math> to itself under the standard group operation, with can build <math>g, 2g, 3g, \ldots</math> as well as <math>-g, -2g, -3g, \ldots</math>. If we let <cmath>P_k = (3k-2014)g</cmath> then we can observe that collinearity between <math>P_a</math>, <math>P_b</math>, and <math>P_c</math> occurs only if <math>P_a + P_b + P_c = 0</math> (by definition of the group operation), which is equivalent to <math>(3a-2014)g + (3b-2014)g + (3c-2014)g = (3a+3b+3c-3*2014)g = 0</math>, or <math>3a + 3b + 3c = 3*2014</math>, or <math>a + b + c = 2014</math>. We know that all these points <math>P_k</math> exist because <math>3k-2014</math> is never 0 for integer <math>k</math>, so that none of these points need to be point at infinity (the identity element of the group).

Revision as of 18:44, 29 April 2014

Problem

Prove that there exists an infinite set of points \[\ldots,\,\,\,\,P_{-3},\,\,\,\,P_{-2},\,\,\,\,P_{-1},\,\,\,\,P_0,\,\,\,\,P_1,\,\,\,\,P_2,\,\,\,\,P_3,\,\,\,\,\ldots\] in the plane with the following property: For any three distinct integers $a,b,$ and $c$, points $P_a$, $P_b$, and $P_c$ are collinear if and only if $a+b+c=2014$.

Solution

Consider an elliptic curve with a generator $g$, such that $g$ is not a root of $0$. By repeatedly adding $g$ to itself under the standard group operation, with can build $g, 2g, 3g, \ldots$ as well as $-g, -2g, -3g, \ldots$. If we let \[P_k = (3k-2014)g\] then we can observe that collinearity between $P_a$, $P_b$, and $P_c$ occurs only if $P_a + P_b + P_c = 0$ (by definition of the group operation), which is equivalent to $(3a-2014)g + (3b-2014)g + (3c-2014)g = (3a+3b+3c-3*2014)g = 0$, or $3a + 3b + 3c = 3*2014$, or $a + b + c = 2014$. We know that all these points $P_k$ exist because $3k-2014$ is never 0 for integer $k$, so that none of these points need to be point at infinity (the identity element of the group).

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