Difference between revisions of "1976 USAMO Problems/Problem 2"
m (→Solution) |
|||
Line 49: | Line 49: | ||
{{USAMO box|year=1976|num-b=1|num-a=3}} | {{USAMO box|year=1976|num-b=1|num-a=3}} | ||
+ | {{MAA Notice}} | ||
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Revision as of 18:02, 3 July 2013
Problem
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.
Solution
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.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1976 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.