Difference between revisions of "2009 USAMO Problems/Problem 1"
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== Solution == | == Solution == | ||
− | Let <math>\omega_3</math> be the [[circumcircle]] of <math>PQRS</math> | + | Let <math>\omega_3</math> be the [[circumcircle]] of <math>PQRS</math>, <math>r_i</math> to be the radius of <math>\omega_i</math>, and <math>O_i</math> to be the center of the circle <math>\omega_i</math>, where <math>i \in \{1,2,3\}</math>. Note that <math>SR</math> and <math>PQ</math> are the [[radical axis]]es of <math>O_1</math> , <math>O_3</math> and <math>O_2</math> , <math>O_3</math> respectively. Hence, by [[power of a point]](the power of <math>O_1</math> can be expressed using circle <math>\omega_2</math> and <math>\omega_3</math> and the power of <math>O_2</math> can be expressed using circle <math>\omega_1</math> and <math>\omega_3</math>), |
− | <cmath> | + | <cmath>O_1O_2^2 - r_2^2 = O_1O_3^2 - r_3^2</cmath> |
− | Subtracting | + | <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>. | ||
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+ | ~AopsUser101 | ||
== See also == | == See also == |
Revision as of 15:52, 1 August 2019
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
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.