https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Timeroot&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-15T23:47:02Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2014_USAMO_Problems/Problem_3&diff=61811 2014 USAMO Problems/Problem 3 2014-04-29T23:44:10Z <p>Timeroot: Added elliptic curve solution</p> <hr /> <div>==Problem==<br /> Prove that there exists an infinite set of points &lt;cmath&gt;\ldots,\,\,\,\,P_{-3},\,\,\,\,P_{-2},\,\,\,\,P_{-1},\,\,\,\,P_0,\,\,\,\,P_1,\,\,\,\,P_2,\,\,\,\,P_3,\,\,\,\,\ldots&lt;/cmath&gt; in the plane with the following property: For any three distinct integers &lt;math&gt;a,b,&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt;, points &lt;math&gt;P_a&lt;/math&gt;, &lt;math&gt;P_b&lt;/math&gt;, and &lt;math&gt;P_c&lt;/math&gt; are collinear if and only if &lt;math&gt;a+b+c=2014&lt;/math&gt;.<br /> <br /> ==Solution==<br /> <br /> Consider an elliptic curve with a generator &lt;math&gt;g&lt;/math&gt;, such that &lt;math&gt;g&lt;/math&gt; is not a root of &lt;math&gt;0&lt;/math&gt;. By repeatedly adding &lt;math&gt;g&lt;/math&gt; to itself under the standard group operation, with can build &lt;math&gt;g, 2g, 3g, \ldots&lt;/math&gt; as well as &lt;math&gt;-g, -2g, -3g, \ldots&lt;/math&gt;. If we let &lt;cmath&gt;P_k = (3k-2014)g&lt;/cmath&gt; then we can observe that collinearity between &lt;math&gt;P_a&lt;/math&gt;, &lt;math&gt;P_b&lt;/math&gt;, and &lt;math&gt;P_c&lt;/math&gt; occurs only if &lt;math&gt;P_a + P_b + P_c = 0&lt;/math&gt; (by definition of the group operation), which is equivalent to &lt;math&gt;(3a-2014)g + (3b-2014)g + (3c-2014)g = (3a+3b+3c-3*2014)g = 0&lt;/math&gt;, or &lt;math&gt;3a + 3b + 3c = 3*2014&lt;/math&gt;, or &lt;math&gt;a + b + c = 2014&lt;/math&gt;. We know that all these points &lt;math&gt;P_k&lt;/math&gt; exist because &lt;math&gt;3k-2014&lt;/math&gt; is never 0 for integer &lt;math&gt;k&lt;/math&gt;, so that none of these points need to be point at infinity (the identity element of the group).</div> Timeroot https://artofproblemsolving.com/wiki/index.php?title=2012_USAJMO_Problems/Problem_2&diff=46603 2012 USAJMO Problems/Problem 2 2012-04-25T15:29:43Z <p>Timeroot: Redirected page to 2012 USAMO Problems/Problem 1</p> <hr /> <div>#REDIRECT [[2012 USAMO Problems/Problem 1]]</div> Timeroot https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_8&diff=36836 1984 AIME Problems/Problem 8 2011-02-11T03:41:08Z <p>Timeroot: </p> <hr /> <div>== Problem ==<br /> The equation &lt;math&gt;z^6+z^3+1&lt;/math&gt; has [[complex root]]s with argument &lt;math&gt;\theta&lt;/math&gt; between &lt;math&gt;90^\circ&lt;/math&gt; and &lt;math&gt;180^\circ&lt;/math&gt; in the [[complex plane]]. Determine the degree measure of &lt;math&gt;\theta&lt;/math&gt;.<br /> <br /> == Solution 1 ==<br /> If &lt;math&gt;r&lt;/math&gt; is a root of &lt;math&gt;z^6+z^3+1&lt;/math&gt;, then &lt;math&gt;0=(r^3-1)(r^6+r^3+1)=r^9-1&lt;/math&gt;. The polynomial &lt;math&gt;x^9-1&lt;/math&gt; has all of its roots with [[absolute value]] &lt;math&gt;1&lt;/math&gt; and argument of the form &lt;math&gt;40m^\circ&lt;/math&gt; for integer &lt;math&gt;m&lt;/math&gt;.<br /> <br /> This reduces &lt;math&gt;\theta&lt;/math&gt; to either &lt;math&gt;120^{\circ}&lt;/math&gt; or &lt;math&gt;160^{\circ}&lt;/math&gt;. But &lt;math&gt;\theta&lt;/math&gt; can't be &lt;math&gt;120^{\circ}&lt;/math&gt; because if &lt;math&gt;r=\cos 120^\circ +i\sin 120^\circ &lt;/math&gt;, then &lt;math&gt;r^3=1&lt;/math&gt; and &lt;math&gt;r^6+r^3+1=3&lt;/math&gt;, a contradiction. This leaves &lt;math&gt;\boxed{\theta=160}&lt;/math&gt;.<br /> <br /> Also,<br /> <br /> From above, you notice that &lt;math&gt;z^6+z^3+1 = \frac {r^9-1}{r^3-1}&lt;/math&gt;. Therefore, the solutions are all of the ninth roots of unity that are not the third roots of unity. After checking, the only angle is &lt;math&gt;\boxed{\theta=160}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Note that the substitution &lt;math&gt;y=z^3&lt;/math&gt; simplifies this to &lt;math&gt;y^2+y+1&lt;/math&gt;. Simply applying the quadratic formula gives roots &lt;math&gt;y_{1,2}=\frac{1}{2}\pm \frac{\sqrt{3}i}{2}&lt;/math&gt;, which have angles of 120 and 240, respectively. This means &lt;math&gt;arg(z) = \frac{120,240}{3} + \frac{360n}{3}&lt;/math&gt;, and the only one between 90 and 180 is &lt;math&gt;\boxed{\theta=160}&lt;/math&gt;.<br /> == See also ==<br /> {{AIME box|year=1984|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Trigonometry Problems]]</div> Timeroot