1976 USAMO Problems/Problem 2
If and are fixed points on a given circle and is a variable diameter of the same circle, determine the locus of the point of intersection of lines and . You may assume that is not a diameter.
WLOG, assume that the circle is the unit circle centered at the origin. Then the points and have coordinates and respectively and and have coordinates and . Note that these coordinates satisfy and since these points are on a unit circle. Now we can find equations for the lines: Solving these simultaneous equations gives coordinates for in terms of and : . These coordinates can be parametrized in Cartesian variables as follows: Now solve for and to get and . Then since which reduces to This equation defines a circle and is the locus of all intersection points . In order to define this locus more generally, find the slope of this circle function using implicit differentiation: Now note that at points and , this slope expression reduces to and respectively, values which are identical to the slopes of lines and . Thus we conclude that the complete locus of intersection points is the circle tangent to lines and at points and respectively.
Notice that (Inscried angle theorem) and that since is a diameter, and thus subtends an arc of . This will hold for all and all , and so by AA similarity, the angle will be constant for all P, thus implying that the points A, B, and all P will be concyclic. If we assume that the center of this circle is , we know that . We can assume that and intersect the original circle at points , and respectively. This will give us that (Since lies on the perpendicular bisector of , we know that , will be on the same side of .) Now we also know that or . The only case where satisfies the measure of , is when and , implying that and are tangents, and so is the intersection of the tangents from and to the original circle.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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