Library / All-Soviet-Union Mathematical Olympiad /167 of 185

All-Soviet-Union Mathematical Olympiad

Geometry Difficulty 6.3 National Olympiad Prove it Soviet Union

Problem:
A circle is circumscribed about the triangle $ABC$ . $X$ is the midpoint of the arc $BC$ (on the opposite side of $BC$ to $A$ ), $Y$ is the midpoint of the arc $AC$ , and $Z$ is the midpoint of the arc $AB$ . $YZ$ meets $AB$ at $D$ and $YX$ meets $BC$ at $E$ . Prove that $DE$ is parallel to $AC$ and that $DE$ passes through the center of the inscribed circle of $ABC$ .

Solution

Solution:
$ZY$ bisects the angle $AYB$ , so $AD/BD = AY/BY$ . Similarly, $XY$ bisects angle $BYC$ , so $CE/BE = CY/BY$ . But $AY = CY$ . Hence $AD/BD = CE/BE$ . Hence triangles $BDE$ and $BAC$ are similar and $DE$ is parallel to $AC$ .

Let $BY$ intersect $AC$ at $W$ and $AX$ at $I$ . $I$ is the incenter. $AI$ bisects angle $BAW$ , so $WI/IB = AW/AB$ . Now consider the triangles $AYW$ , $BYA$ . Clearly $\angle AYW = \angle BYA$ . Also $\angle WAY = \angle CAY = \angle ABY$ . Hence the triangles are similar and $AW/AY = AB/BY$ . So $AW/AB = AY/BY$ . Hence $WI/IB = AY/BY = AD/BD$ . So triangles $BDI$ and $BAW$ are similar and $DI$ is parallel to $AW$ and hence to $DE$ . So $DE$ passes through $I$ .

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.