2009 USAMO Problems/Problem 1
Contents
Problem
Given circles and intersecting at points and , let be a line through the center of intersecting at points and and let be a line through the center of intersecting at points and . Prove that if and lie on a circle then the center of this circle lies on line .
Solution 1
Let be the circumcircle of , to be the radius of , and to be the center of the circle , where . Note that and are the radical axises of , and , respectively. Hence, by power of a point(the power of can be expressed using circle and and the power of can be expressed using circle and ), Subtracting these two equations yields that , so must lie on the radical axis of , .
~AopsUser101
Solution 2
Define and similarly to above. Note that is perpendicular to RS and is perpendicular to PQ. Thus, the intersection of PQ and RS must be the orthocenter of triangle . Define this as point . Extending line to meet , we note that is perpendicular to .
In addition, note that by the radical axis theorem, the intersection of and must also lie on the radical axis of and . Because the radical axis of and is perpendicular to and contains , it must also contain , and we are done.
See also
2009 USAMO (Problems • Resources) | ||
Preceded by First question |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
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