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Winter Mathematical Competition

Geometry Difficulty 6.4 National Olympiad Prove it Bulgaria

In $\triangle ABC$ with $\angle ACB = 60^\circ$ let $AA_1$ and $BB_1$ ( $A_1 \in BC$ , $B_1 \in AC$ ) be the bisectors of $\angle BAC$ and $\angle ABC$ . The line $A_1B_1$ meets the circumcircle of $\triangle ABC$ at points $A_2$ and $B_2$ .

a) If $O$ and $I$ are the circumcenter and the incenter of $\triangle ABC$ prove that $OI$ is parallel to $A_1B_1$ .

b) If $R$ is the midpoint of the arc $\widehat{AB}$ , not containing point $C$ , and $P$ and $Q$ are the midpoints of $A_1B_1$ and $A_2B_2$ , respectively, prove that $RP = RQ$ .

Solution

a) Since $\angle AOB = 2\gamma = 120^\circ$ and points $A$ , $O$ , $I$ and $B$ lie on a circle
$\alpha + \beta$

b) Since $OQ \perp A_2B_2$ , $IP \perp A_2B_2$ ( $\triangle A_1IB_1$ is isosceles) and $OI \parallel A_2B_2$ , we have that $OIPQ$ is a rectangle. The perpendicular bisector of $OI$ is also the perpendicular bisector of $PQ$ . Using that $R$ lies on the perpendicular bisector of $OI$ it follows that $R$ lies on the perpendicular bisector of $PQ$ , i.e. $RP = RQ$ .

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