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PrintBalkan 2012 shortlist
2012 geometry
Problem
Let be the point of intersection of the diagonals of a cyclic quadrilateral . Let and be the incenters of triangles and , respectively, and let be the point of intersection of the lines and . The foot of the perpendicular from the midpoint of to is , and is the midpoint of . Let and be the points of intersection of the line with and , respectively. Let be the circumcenter of triangle , and let and be the circles with diameters and , respectively. Let and be the second points of intersection of with and , respectively. If is the point where the circles and meet again, prove that is the circumcenter of the triangle .
Solution
The point is the midpoint of the arc and, as , , are collinear, we have . Therefore the triangle is isosceles and .
Let be the midpoint of . Then , so and it follows that is the orthocenter of the triangle . Therefore we have .
Let be the orthogonal projection of to and be the other end point of diameter of through . Since and are diameters, and . Since is cyclic, . As we also have and , the triangles and are congruent and .
On the other hand, since the quadrilaterals and are cyclic, we have so the triangle is isosceles and . Hence is the circumcenter of the triangle .
Let be the midpoint of . Then , so and it follows that is the orthocenter of the triangle . Therefore we have .
Let be the orthogonal projection of to and be the other end point of diameter of through . Since and are diameters, and . Since is cyclic, . As we also have and , the triangles and are congruent and .
On the other hand, since the quadrilaterals and are cyclic, we have so the triangle is isosceles and . Hence is the circumcenter of the triangle .
Techniques
Cyclic quadrilateralsTriangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circleAngle chasing