Difference between revisions of "2009 USAMO Problems/Problem 1"
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Define <math>\omega_i</math> and <math>O_i</math> similarly to above. Note that <math>O_1O_3</math> is perpendicular to <math>RS</math> and <math>O_2 O_3</math> is perpendicular to <math>PQ</math>. Thus, the intersection of <math>PQ</math> and <math>RS</math> 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>. | Define <math>\omega_i</math> and <math>O_i</math> similarly to above. Note that <math>O_1O_3</math> is perpendicular to <math>RS</math> and <math>O_2 O_3</math> is perpendicular to <math>PQ</math>. Thus, the intersection of <math>PQ</math> and <math>RS</math> 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 | + | 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. |
− | + | It seems like at least but the radical axis theorem has its own conditions. Consider a very special when all centers are collinear. | |
== See also == | == See also == |
Revision as of 08:35, 15 April 2021
Contents
[hide]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
and
is perpendicular to
. Thus, the intersection of
and
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.
It seems like at least but the radical axis theorem has its own conditions. Consider a very special when all centers are collinear.
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.