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 .
