Difference between revisions of "2009 USAMO Problems/Problem 1"
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== Solution 2 == | == Solution 2 == | ||
− | Define <math>\omega_i</math> and <math>O_i</math> similarly to above. Note that | + | Define <math>\omega_i</math> and <math>O_i</math> similarly to above. Note that <math>O_1O_3</math> is perpendicular to RS and <math>O_2 O_3</math> is perpendicular to PQ. Thus, the intersection of PQ and RS must be the orthocenter of triangle <math>O_1O_2O_3</math>. Define this as point <math>H</math>. Extending line <math>O_3H</math> to meet <math>O_1O_2</math>, we note that <math>O_3H</math> is perpendicular to <math>O_1O_2</math>. |
+ | |||
+ | In addition, note that by the radical axis theorem, the intersection of <math>PQ</math> and <math>RS</math> must also lie on the radical axis of <math>\omega_1</math> and <math>\omega_2</math>. Because the radical axis of <math>\omega_1</math> and <math>\omega_2</math> is perpendicular to <math>O_1O_2</math> and contains <math>H</math>, it must also contain <math>O_3</math>, and we are done. | ||
== See also == | == See also == |
Revision as of 20:24, 8 August 2020
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.