Difference between revisions of "2009 USAMO Problems/Problem 1"
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<cmath> O_2O_1^2 - r_1^2 = O_2O_3^2 - r_3^2</cmath> | <cmath> O_2O_1^2 - r_1^2 = O_2O_3^2 - r_3^2</cmath> | ||
Subtracting these two equations yields that <math>O_1O_3^2 - r_1^2 = O_2O_3^2 - r_2^2</math>, so <math>O_3</math> must lie on the [[radical axis]] of <math>\omega_1</math> , <math>\omega_2</math>. | Subtracting these two equations yields that <math>O_1O_3^2 - r_1^2 = O_2O_3^2 - r_2^2</math>, so <math>O_3</math> must lie on the [[radical axis]] of <math>\omega_1</math> , <math>\omega_2</math>. | ||
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==Remarks== | ==Remarks== |
Revision as of 14:54, 1 August 2019
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
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
Remarks
Conveniently, this analytic solution takes care of all configuration issues we may have encountered had we used a more traditional solution, such as angle-chasing(which would indeed work).
~AopsUser101
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.