Difference between revisions of "2011 USAJMO Problems/Problem 3"
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By Lemma 2, it follows that the reflections of <math>F</math> across <math>\overleftrightarrow{XY}</math> and <math>\overleftrightarrow{XZ}</math>, call them <math>P_2'</math> and <math>P_3'</math>, lie on <math>d</math>. Let the intersection of <math>\overleftrightarrow{YZ}</math> and the perpendicular to <math>d</math> through <math>P_1'</math> be <math>P_1</math>, the intersection of <math>\overleftrightarriw{XY}</math> and the perpendicular to <math>d</math> through <math>P_2'</math> be <math>P_2</math>, and the intersection of <math>\overleftrightarrow{XZ}</math> and the perpendicular to <math>d</math> through <math>P_3'</math> be <math>P_3</math>. Then by the definitions of <math>P_1'</math>, <math>P_2'</math>, and <math>P_3'</math> it follows that <math>FP_i=P_iP_i'</math> for <math>i=1,2,3</math> and so <math>P_1</math>, <math>P_2</math>, and <math>P_3</math> are on <math>p</math>. By lemma 1, <math>\ell(P_1)=\overleftrightarrow{YZ}</math>, <math>\ell(P_2)=\overleftrightarrow{XY}</math>, and <math>\ell(P_3)=\overleftrightarrow{XZ}</math>. Therefore the intersections of <math>\ell(P_1)</math>, <math>\ell(P_2)</math>, and <math>\ell(P_3)</math> form an equilateral triangle with center <math>H</math>, which finishes the proof. | By Lemma 2, it follows that the reflections of <math>F</math> across <math>\overleftrightarrow{XY}</math> and <math>\overleftrightarrow{XZ}</math>, call them <math>P_2'</math> and <math>P_3'</math>, lie on <math>d</math>. Let the intersection of <math>\overleftrightarrow{YZ}</math> and the perpendicular to <math>d</math> through <math>P_1'</math> be <math>P_1</math>, the intersection of <math>\overleftrightarriw{XY}</math> and the perpendicular to <math>d</math> through <math>P_2'</math> be <math>P_2</math>, and the intersection of <math>\overleftrightarrow{XZ}</math> and the perpendicular to <math>d</math> through <math>P_3'</math> be <math>P_3</math>. Then by the definitions of <math>P_1'</math>, <math>P_2'</math>, and <math>P_3'</math> it follows that <math>FP_i=P_iP_i'</math> for <math>i=1,2,3</math> and so <math>P_1</math>, <math>P_2</math>, and <math>P_3</math> are on <math>p</math>. By lemma 1, <math>\ell(P_1)=\overleftrightarrow{YZ}</math>, <math>\ell(P_2)=\overleftrightarrow{XY}</math>, and <math>\ell(P_3)=\overleftrightarrow{XZ}</math>. Therefore the intersections of <math>\ell(P_1)</math>, <math>\ell(P_2)</math>, and <math>\ell(P_3)</math> form an equilateral triangle with center <math>H</math>, which finishes the proof. | ||
--Killbilledtoucan | --Killbilledtoucan | ||
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Revision as of 12:20, 4 July 2013
Problem
For a point in the coordinate plane, let denote the line passing through with slope . Consider the set of triangles with vertices of the form , , , such that the intersections of the lines , , form an equilateral triangle . Find the locus of the center of as ranges over all such triangles.
Solution
Note that all the points belong to the parabola which we will denote . This parabola has a focus and directrix which we will denote . We will prove that the desired locus is .
First note that for any point on , the line is the tangent line to at . This is because contains and because . If you don't like calculus, you can also verify that has equation and does not intersect at any point besides . Now for any point on let be the foot of the perpendicular from onto . Then by the definition of parabolas, . Let be the perpendicular bisector of $\oveline{P'F}$ (Error compiling LaTeX. Unknown error_msg). Since , passes through . Suppose is any other point on and let be the foot of the perpendicular from to . Then in right , is a leg and so . Therefore cannot be on . This implies that is exactly the tangent line to at , that is . So we have proved Lemma 1: If is a point on then is the perpendicular bisector of .
We need another lemma before we proceed. Lemma 2: If is on the circumcircle of with orthocenter , then the reflections of across , , and are collinear with .
Proof of Lemma 2: Say the reflections of and across are and , and the reflections of and across are and . Then we angle chase where is the measure of minor arc on the circumcircle of . This implies that is on the circumcircle of , and similarly is on the circumcircle of . Therefore , and . So . Since , , and are collinear it follows that , and are collinear. Similarly, the reflection of over also lies on this line, and so the claim is proved.
Now suppose , , and are three points of and let , , and . Also let , , and be the midpoints of , , and respectively. Then since and , it follows that , , and are collinear. By Lemma 1, we know that , , and are the feet of the altitudes from to , , and . Therefore by the Simson Line Theorem, is on the circumcircle of . If is the orthocenter of , then by Lemma 2, it follows that is on . It follows that the locus described in the problem is a subset of .
Since we claim that the locus described in the problem is , we still need to show that for any choice of on there exists an equilateral triangle with center such that the lines containing the sides of the triangle are tangent to . So suppose is any point on and let the circle centered at through be . Then suppose is one of the intersections of with . Let , and construct the ray through on the same halfplane of as that makes an angle of with . Say this ray intersects in a point besides , and let be the perpendicular bisector of . Since and , we have . By the inscribed angles theorem, it follows that . Also since and are both radii, is isosceles and . Let be the reflection of across . Then , and so . It follows that is on , which means is the perpendicular bisector of .
Let intersect in points and and let be the point diametrically opposite to on . Also let intersect at . Then . Therefore is a right triangle and so . So and by the inscribed angles theorem, . Since it follows that is and equilateral triangle with center .
By Lemma 2, it follows that the reflections of across and , call them and , lie on . Let the intersection of and the perpendicular to through be , the intersection of $\overleftrightarriw{XY}$ (Error compiling LaTeX. Unknown error_msg) and the perpendicular to through be , and the intersection of and the perpendicular to through be . Then by the definitions of , , and it follows that for and so , , and are on . By lemma 1, , , and . Therefore the intersections of , , and form an equilateral triangle with center , which finishes the proof. --Killbilledtoucan The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.