2003 JBMO Problems/Problem 3
Let , , be the midpoints of the arcs , , on the circumcircle of a triangle not containing the points , , , respectively. Let the line meet and at and , and let be the midpoint of the segment . Let the line meet and at and , and let be the midpoint of the segment .
a) Find the angles of triangle ;
b) Prove that if is the point of intersection of the lines and , then the circumcenter of triangle lies on the circumcircle of triangle .
Let , intersect , at , respectively. We will prove first that and that lines , , are altitudes of the .
It's easy to see that lines , and form the internal angle bisectors of .
Consequently, we can determine the of as being equal to
Also we have , thus . Similarly .
Thus , , are altitudes of the with , , respectively being the feet of the altitudes.
Now since is internal bisector of and is perpendicular to , we have that is the perpendicular bisector of . Hence .
Similarly it can be shown that is the perpendicular bisector of , and hence .
Now lines , and intersect at point . So is the incenter of and orthocenter of .
Clearly, is a cyclic quadrilateral as , are the feet of perpendiculars from and .
So, we have .
Similarly, since is also a cyclic-quadrilateral, reasoning as above, .
Thus we have that and so is an internal bisector of . Reasoning in a similar fashion it can be proven that and are internal bisectors of other 2 angles of .
Thus also happens to be the incenter of in addition to being that of .
: Angles of :
Since . Similarly . Finally .
Let circumcircle of cut line at point . Since is a cyclic quadrilateral, we have .
Similarly, . Thus = .
and . Thus = .
Thus we have = = . So is the circumcenter of .