Bariz - Problem

Source: Cyprus 2022 Junior TST-2 Problem 3

Tags: geometry, isosceles, circumcircle

Demetres

21.02.2022 15:38

Let $ABC$ be an acute-angled triangle, and let $D, E$ and $K$ be the midpoints of its sides $AB, AC$ and $BC$ respectively. Let $O$ be the circumcentre of triangle $ABC$, and let $M$ be the foot of the perpendicular from $A$ on the line $BC$. From the midpoint $P$ of $OM$ we draw a line parallel to $AM$, which meets the lines $DE$ and $OA$ at the points $T$ and $Z$ respectively. Prove that: (a) the triangle $DZE$ is isosceles (b) the area of the triangle $DZE$ is given by the formula \[E_{DZE}=\frac{BC\cdot OK}{8}\]

Statement is much scarier than the problem itself. a) Just note that $DMEK$ is isosceles trapezoid, since $D,M,E,K$ are concyclic and $\overline{MK}\parallel \overline{DE}$. Thus, perpendicular bisectors of $\overline{DE},\overline{MK}$ coincide. Therefore, $Z$ is the center of $(ADEO)$ as $D$ lies on $\overline{AO}$ and perpendicular bisector of $\overline{DE}$. We conclude that $\triangle DZE$ is isosceles. b) By homothety, $\triangle DZE\sim\triangle BOC$ with the ratio of similarity $2$. Thus, $$E_{DZE}=\frac{DE\cdot TZ}{2}=\frac{\tfrac{BC}{2}\cdot \tfrac{OK}{2}}{2}=\frac{BC\cdot OK}{2}.$$