Difference between revisions of "2009 OIM Problems/Problem 4"
(Created page with "== Problem == Let <math>ABC</math> be a triangle with <math>AB \ne AC</math>. Let <math>I</math> be the incenter of <math>ABC</math> and <math>P</math> the other point of inte...") |
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== Solution == | == Solution == | ||
− | + | Clearly <math>P</math> is the midpoint of arc <math>BAC</math>. Let <math>BI</math>, <math>CI</math> intersect the circumcircle of <math>ABC</math> at <math>D</math>, <math>E</math> respectively. It is well known that <math>PDIE</math> is a parallelogram. Therefore, <math>\angle ICJ=\angle ECJ=\angle EPJ=\angle BEJ</math>, which implies BI tangent to the circumcircle of <math>JIC</math>. Similarly, <math>CI</math> is tangent to the circumcircle of <math>JIB</math>. | |
== See also == | == See also == | ||
[[OIM Problems and Solutions]] | [[OIM Problems and Solutions]] |
Latest revision as of 04:59, 26 March 2024
Problem
Let be a triangle with
. Let
be the incenter of
and
the other point of intersection of the exterior bisector of angle
with the circumcircle of
. The line
intersects for the second time the circumcircle of
at point
. Show that the circumcircles of triangles
and
are tangent to
and
, respectively.
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
Clearly is the midpoint of arc
. Let
,
intersect the circumcircle of
at
,
respectively. It is well known that
is a parallelogram. Therefore,
, which implies BI tangent to the circumcircle of
. Similarly,
is tangent to the circumcircle of
.