## Task B-2.7.

Inside the square $A B C D$, there are points $E, F, G$ and $H$ such that the triangles $A B E$, $B C F$, $C D G$ and $D A H$ are equilateral. What is the ratio of the area of the square $A B C D$ to the area of the quadrilateral $E F G H$?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_ab4e7c3a83d03c059773g-11.jpg?height=525&width=532&top_left_y=1342&top_left_x=682)

Points $E$ and $G$ are vertices of equilateral triangles with bases $\overline{A B}$ and $\overline{C D}$, so they lie on the perpendicular bisectors of these segments. Similarly, points $F$ and $H$ lie on the perpendicular bisectors of segments $\overline{B C}$ and $\overline{A D}$.

Since these perpendicular bisectors are mutually perpendicular, triangles $O F E, O G F, O H G$, and $O E H$ are congruent right triangles.

Let the side length of the given square be $a$.

Then $|P G|=a-\frac{a \sqrt{3}}{2}=a\left(1-\frac{\sqrt{3}}{2}\right)$,

$|G O|=\frac{a}{2}-|P G|=a \cdot \frac{\sqrt{3}-1}{2}$

1 point

and similarly $|F O|=|E O|=|H O|=|G O|=a \cdot \frac{\sqrt{3}-1}{2}$.

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Triangles $O F E, O G F, O H G$, and $O E H$ are congruent right triangles, and quadrilateral $E F G H$ is a square.

Its area is given by $P=\frac{1}{2} d^{2}$, where $d=2|F O|$ is the length of the diagonal of this square.

$P=\frac{1}{2}\left(2 \cdot a \cdot \frac{\sqrt{3}-1}{2}\right)^{2}=a^{2}(2-\sqrt{3})$.

The desired ratio of areas is $\frac{P(A B C D)}{P(E F G H)}=\frac{a^{2}}{a^{2}(2-\sqrt{3})}=\frac{1}{2-\sqrt{3}}=2+\sqrt{3}$.

Note: If a student calculates the ratio $\frac{P(E F G H)}{P(A B C D)}=2-\sqrt{3}$ instead of $\frac{P(A B C D)}{P(E F G H)}$, they should receive full marks.

## SCHOOL/CITY COMPETITION IN MATHEMATICS

3rd grade - high school - B variant
January 4, 2020.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.