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>. | ||
+ | |||
+ | ~AopsUser101 | ||
== See also == | == See also == |
Revision as of 14:52, 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.