Library / Problems of Ukrainian Authors /16 of 40

Problems of Ukrainian Authors

Geometry Difficulty 5.4 AIME, harder Prove it Ukraine

On a plane are placed two triangles $ABC$ and $BKL$ such that segment $AK$ is divided into three equal parts by the intersection point of the medians of triangle $ABC$ and the intersection point of the bisectors of triangle $BKL$ ( $AK$ is a median of $ABC$ , $KA$ is a bisector of $ABKL$ ), and quadrilateral $KALC$ is a trapezium. Find all angles of triangle $BKL$ .

Solution

Let $I$ be the center of $ABKL$ , contextual from the characteristic of the centroid of the triangle and from the condition of the task: $AI = x$ , $KI = 2x$ . Let $BA = x$ , then from the characteristic of the bisector $BK = 2x$ and $KC = 2x$ . Then $KALC$ is a trapezium. That's why $KA \parallel LC$ (see Fig. 20). From Thales' theorem $AL = x$ . Therefore, in $ABKL$ , $KA$ is a bisector by condition and a median because $BA = AL$ . Then $ABKL$ is an equilateral triangle, and so if $BK = 2x = BL$ , therefore it is an equilateral triangle and all its angles are $60^\circ$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.