2014 USAMO Problems/Problem 3

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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2014 USAMO Problems/Problem 3

Problem

Solution