$\left[\begin{array}{l}\text { [Lengths of sides, heights, medians, and angle bisectors] } \\ \text { [ }\end{array}\right.$

In an isosceles triangle $ABC$ with base $AC$, point $D$ divides side $BC$ in the ratio $3:1$, counting from vertex $B$, and point $E$ is the midpoint of segment $AD$. It is known that $BE=\sqrt{7}$ and $CE=3$. Find the radius of the circumcircle of triangle $ABC$.
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 $C D=x$. Then $B D=3 x, A B=B C=4 x$. If $M$ is the midpoint of side $B C$, then $E M$ is the midline of triangle $A B D$, so $E M=2 x$. On the other hand, $E M$ is the median of triangle $B E C$, so

$$
E M^{2}=\frac{1}{4}\left(2 B E^{2}+2 C E^{2}-B C^{2}\right), \text { or } 4 x^{2}=2 \cdot 7+2 \cdot 9-16 x^{2} \text {, }
$$

from which we find that $x=1$.

Since $D E$ is the median of triangle $M E C$, then

$$
E D^{2}=\frac{1}{4}\left(2 M E^{2}+2 C E^{2}-M C^{2}\right)=\frac{1}{4}\left(2 \cdot 4 x^{2}+2 \cdot 9-4 x^{2}\right)=\frac{22}{4}
$$

Thus,

$$
E D=\frac{\sqrt{22}}{2}, A D=\sqrt{22}
$$

Since $C E$ is the median of triangle $A C D$, then

$$
C E^{2}=\frac{1}{4}\left(2 A C^{2}+2 C D^{2}-A D^{2}\right), \text { or } 9=\frac{1}{4}\left(2 A C^{2}+2-22\right),
$$

from which we find that $A C=\sqrt{28}$. Then

$$
\cos \angle A C B=\frac{A C}{2 B C}=\frac{\sqrt{28}}{8}=\frac{\sqrt{7}}{4}, \sin \angle A C B=\frac{3}{4} .
$$

If $R$ is the radius of the circumscribed circle of triangle $A B C$, then

$$
R=\frac{A B}{2 \sin \angle A C B}=\frac{4}{\frac{3}{2}}=\frac{8}{3} \text {. }
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-38.jpg?height=412&width=717&top_left_y=154&top_left_x=676)

## Answer

$\frac{8}{3}$.

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