## [ Hexagons ] [ Midline of a trapezoid ] Difficult [Ratio of areas of triangles with a common base or common height] Classes [ The area of a figure is equal to the sum of the areas of the figures into which it is divided ]

Authors: Lopshits A.L., Vasilyev N.B.

Let $A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}, E^{\prime}, F^{\prime}$ be the midpoints of the sides $A B, B C, C D, D E, E F, F A$ of an arbitrary convex hexagon $A B C D E F$. The areas of triangles $A B C^{\prime}, B C D^{\prime}, C D E^{\prime}, D E F^{\prime}, E F A^{\prime}, F A B^{\prime}$ are known. Find the area of the hexagon $A B C D E F$.
Drop perpendiculars $C K, C^{\prime} L, D M$ from points $C, C^{\prime}, D$ to $A B$. Note that $C^{\prime} L=1 / 2(C K+D M)$ as the midline of trapezoid $K C D M$. Considering triangles $A B C, A B C^{\prime}, A B D$ with the common base $A B$ and heights $C K, C^{\prime} L, D M$ respectively, we obtain the relation $2 S_{A B C^{\prime}}=S_{A B C}+S_{A B D}$. Write similar relations for all the triangles mentioned in the problem and sum them up:

$2 \cdot\left(S_{A B C^{\prime}}+S_{B C D^{\prime}}+S_{C D E^{\prime}}+S_{D E F^{\prime}}+S_{E F A^{\prime}}+S_{F A B^{\prime}}\right)=S_{A B C}+S_{A B D}+S_{B C D}+S_{B C E}+S_{C D E}+S_{C D F}+S_{D E F}+S_{D E A}$ $+S_{E F A}+S_{E F B}+S_{F A B}+S_{F A C}=$

$=\left(S_{A B C}+S_{F A C}+S_{C D F}+S_{D E F}\right)+\left(S_{B C D}+S_{A B D}+S_{D E A}+S_{E F A}\right)+\left(S_{C D E}+S_{B C E}+S_{E F B}+S_{F A B}\right)=3 S_{A B C D E F}$ (triangles in each parenthesis completely cover the hexagon).

## Answer

$2 / 3$ of the sum of the areas of the specified triangles.