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Estonia — Open Contests

Geometry Difficulty 5.0 AIME, harder Prove it Estonia

Let $A$ and $B$ be such points of the circle with centre $O$ that the triangle $AOB$ is right-angled. The perpendicular bisector of the segment $AO$ intersects the shorter arc $AB$ in point $K$ . The lines $KO$ and $AB$ intersect in point $L$ . Prove that the triangle $KBL$ is isosceles.

Solution

Fig. 4

Since $K$ lies on the perpendicular bisector of the segment $AO$ (Fig. 4) we have $KA = KO$ . On the other hand $KO = AO$ since $K$ and $A$ are points on the circle. Hence $AKO$ is an equilateral triangle from which we obtain that $\angle AOK = 60^\circ$ . Hence $\angle KOB = \angle AOB - \angle AOK = 90^\circ - 60^\circ = 30^\circ$ . As also $B$ lies on the same circle we have $KO = BO$ , hence $\angle BKO = \frac{180^\circ - \angle KOB}{2} = \frac{180^\circ - 30^\circ}{2} = 75^\circ$ . Finally from the equality $AO = BO$ we obtain $\angle ABO = \frac{180^\circ - \angle AOB}{2} = \frac{90^\circ}{2} = 45^\circ$ . Hence $\angle BLK = \angle LOB + \angle LBO = 30^\circ + 45^\circ = 75^\circ$ . The triangle $KBL$ is isosceles as it has two equal angles.

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