AoPS - 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 $E_{DZE}=\frac{BC\cdot OK}{8}$

rafaello

21.02.2022 17:00

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, $E_{DZE}=\frac{DE\cdot TZ}{2}=\frac{\tfrac{BC}{2}\cdot \tfrac{OK}{2}}{2}=\frac{BC\cdot OK}{2}.$

rafigamath

04.06.2023 14:19

Since $ZP \parallel AM$ and $OP=PM$ $\implies AZ=ZO$ . And $AD=DB$ , $AE=EC$ $\to DZ\parallel OB$ and $ZE\parallel OC$ . From Thales, we have $2DZ=OB , 2ZE=OC$ $\to DZ=ZE$ . And $E_{DZE}=\frac{DE\cdot TZ}{2}=\frac{\tfrac{BC}{2}\cdot \tfrac{OK}{2}}{2}=\frac{BC\cdot OK}{2}.$ and this finishes the problem

wizixez

24.12.2024 20:10

Lol..can finish proof by using elementary complex bash info