8,9

Given a convex quadrilateral of area S. Inside it, a point is chosen and reflected symmetrically relative to the midpoints of its sides. Four vertices of a new quadrilateral are obtained. Find its area.
The area of a quadrilateral with vertices at the midpoints of the sides of a given convex quadrilateral is half the area of the given quadrilateral.

## Solution

Let $M, N, K$, and $L$ be the midpoints of the sides $AB, BC, CD$, and $AD$ of the quadrilateral $ABCD$, respectively, and $P$ be a point inside this quadrilateral; $X, Y, Z$, and $T$ be the images of point $P$ under symmetry about points $M, N, K$, and $L$, respectively.

Then $MN$ is the midline of triangle $XPY$. Therefore,

$$
S_{\Delta \mathrm{XPY}}=\frac{1}{4} S_{\Delta \mathrm{MNP}}
$$

Writing similar equalities for triangles $YPZ, ZPT$, and $TPX$, we get that

$$
S_{\mathrm{XYZT}}=4 S_{\mathrm{MNKL}}=4 \cdot \frac{1}{2} S=2 S
$$

Let $M, N, K$, and $L$ be the midpoints of the sides $AB, BC, CD$, and $AD$ of the quadrilateral $ABCD$, respectively, and $P$ be a point inside this quadrilateral; $X, Y, Z$, and $T$ be the images of point $P$ under symmetry about points $M, N, K$, and $L$, respectively.

Then $MN$ is the midline of triangle $XPY$. Therefore,

$$
S_{\Delta \mathrm{XPY}}=\frac{1}{4} S_{\Delta \mathrm{MNP}}
$$

Writing similar equalities for triangles $YPZ, ZPT$, and $TPX$, we get that

$$
S_{\mathrm{XYZT}}=4 S_{\mathrm{MNKL}}=4 \cdot \frac{1}{2} S=2 S
$$

Let $M, N, K$, and $L$ be the midpoints of the sides $AB, BC, CD$, and $AD$ of the quadrilateral $ABCD$, respectively, and $P$ be a point inside this quadrilateral; $X, Y, Z$, and $T$ be the images of point $P$ under symmetry about points $M, N, K$, and $L$, respectively.

Then $MN$ is the midline of triangle $XPY$. Therefore,

$$
S_{\Delta \mathrm{XPY}}=\frac{1}{4} S_{\Delta \mathrm{MNP}}
$$

Writing similar equalities for triangles $YPZ, ZPT$, and $TPX$, we get that

$$
S_{\mathrm{XYZT}}=4 S_{\mathrm{MNKL}}=4 \cdot \frac{1}{2} S=2 S
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-21.jpg?height=569&width=778&top_left_y=1282&top_left_x=640)

## Answer

$2 S$.