[Lengths of sides, heights, medians, and bisectors]

In rhombus $A B C D$, point $Q$ divides side $B C$ in the ratio $1: 3$, counting from vertex $B$, and point $E$ is the midpoint of side $A B$. It is known that the median $C F$ of triangle $C E Q$ is $2 \sqrt{2}$, and $E Q=\sqrt{2}$. Find the radius of the circle inscribed in rhombus $A B C D$.
Use the formula for the median of a triangle:

$$
m^{2}=\frac{1}{4}\left(2 a^{2}+2 b^{2}-c^{2}\right)
$$

where $a, b, c$ are the sides of the triangle, and $m$ is the median drawn to the side equal to $c$.

## Solution

We will use the formula for the median of a triangle:

$$
m^{2}=\frac{1}{4}\left(2 a^{2}+2 b^{2}-c^{2}\right)
$$

where $a, b, c$ are the sides of the triangle, and $m$ is the median drawn to the side equal to $c$.

Let $B Q=x, A C=2 y$. Let $O$ be the intersection point of the diagonals of the rhombus $A B C D$, and $M$ be the midpoint of side $B C$. Then

$$
O C=y, A B=B C=4 x, M Q=x
$$

and since $M E$ is the midline of triangle $A B C$, we have $M E=\frac{1}{2} A C=y$.

Since $E Q$ is the median of triangle $B E M$, we have

$$
E Q^{2}=\frac{1}{4}\left(2 B E^{2}+2 E M^{2}-B M^{2}\right) \text {, or } 2=\frac{1}{4}\left(2 \cdot 4 x^{2}+2 y^{2}-4 x^{2}\right)
$$

or

$$
2 x^{2}+y^{2}=4
$$

Since $C E$ is the median of triangle $A B C$, we have

$$
C E^{2}=\frac{1}{4}\left(2 A C^{2}+2 B C^{2}-A B^{2}\right)=\frac{1}{4}\left(8 y^{2}+32 x^{2}-16 x^{2}\right)=4 x^{2}+2 y^{2}
$$

Since $C F$ is the median of triangle $C E Q$, we have

$$
8=C F^{2}=\frac{1}{4}\left(2 C E^{2}+2 C Q^{2}-E Q^{2}\right)=\frac{1}{4}\left(8 x^{2}+4 y^{2}+18 x^{2}-2\right)=\frac{1}{4}\left(26 x^{2}+4 y^{2}-2\right)
$$

from which

$$
13 x^{2}+y^{2}=17
$$

From the obtained system of equations

$$
\left\{\begin{array}{l}
2 x^{2}+y^{2}=4 \\
13 x^{2}+y^{2}=17
\end{array}\right.
$$

we find that $x=1, y=\sqrt{2}$. Then

$$
O B^{2}=B C^{2}-O C^{2}=16-2=14
$$

Let $O P$ be the height of the right triangle $B O C$. Then $O P$ is the radius of the circle inscribed in the rhombus. Since $B C \cdot O P=O B \cdot O C$, we have

$$
O P=\frac{O B \cdot O C}{B C}=\frac{\sqrt{14} \cdot \sqrt{2}}{4}=\frac{\sqrt{7}}{2} \text {. }
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_f7f9d90261bfc416bd71g-13.jpg?height=471&width=734&top_left_y=1794&top_left_x=662)

Answer

$\frac{\sqrt{7}}{2}$.