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PrintWinter Mathematical Competition
Bulgaria geometry
Problem
In with let and (, ) be the bisectors of and . The line meets the circumcircle of at points and .
a) If and are the circumcenter and the incenter of prove that is parallel to .
b) If is the midpoint of the arc , not containing point , and and are the midpoints of and , respectively, prove that .

a) If and are the circumcenter and the incenter of prove that is parallel to .
b) If is the midpoint of the arc , not containing point , and and are the midpoints of and , respectively, prove that .
Solution
a) Since and points , , and lie on a circle
b) Since , ( is isosceles) and , we have that is a rectangle. The perpendicular bisector of is also the perpendicular bisector of . Using that lies on the perpendicular bisector of it follows that lies on the perpendicular bisector of , i.e. .
b) Since , ( is isosceles) and , we have that is a rectangle. The perpendicular bisector of is also the perpendicular bisector of . Using that lies on the perpendicular bisector of it follows that lies on the perpendicular bisector of , i.e. .
Techniques
Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circleCyclic quadrilateralsAngle chasingDistance chasing