Difference between revisions of "2013 USAMO Problems/Problem 6"

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Let <math>ABC</math> be a triangle.  Find all points <math>P</math> on segment <math>BC</math> satisfying the following property:  If <math>X</math> and <math>Y</math> are the intersections of line <math>PA</math> with the common external tangent lines of the circumcircles of triangles <math>PAB</math> and <math>PAC</math>, then <cmath>\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1</cmath>.
 
Let <math>ABC</math> be a triangle.  Find all points <math>P</math> on segment <math>BC</math> satisfying the following property:  If <math>X</math> and <math>Y</math> are the intersections of line <math>PA</math> with the common external tangent lines of the circumcircles of triangles <math>PAB</math> and <math>PAC</math>, then <cmath>\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1</cmath>.
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Revision as of 17:59, 3 July 2013

Let $ABC$ be a triangle. Find all points $P$ on segment $BC$ satisfying the following property: If $X$ and $Y$ are the intersections of line $PA$ with the common external tangent lines of the circumcircles of triangles $PAB$ and $PAC$, then \[\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1\]. The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png