In isosceles triangle $ABC$ with base $AC$, point $D$ divides side $BC$ in the ratio $2:1$, counting from vertex $B$, and point $E$ is the midpoint of side $AB$. It is known that the median $CQ$ of triangle $CED$ is $\frac{\sqrt{23}}{2}$, and $DE = \frac{\sqrt{23}}{2}$. Find the radius of the circumcircle of triangle $ABC$.
Let $A B=B C=6 a, \angle B A C=\angle B C A=\alpha$ and using the cosine theorem and the formula for the median of a triangle, form a system of equations in terms of $a^{2}$ and $\cos ^{2} \alpha$.

## Solution

Let $A B=B C=6 a, \angle B A C=\angle B C A=\alpha$. Then

$$
A E=B E=3 a, B D=4 a, C D=2 a, A C=12 a \cos \alpha, \angle A B C=180^{\circ}-2 \alpha
$$

From triangles $A E C$ and $B D E$ using the cosine theorem, we find that

$$
\begin{gathered}
C E^{2}=9 a^{2}+144 a^{2} \cos ^{2} \alpha-72 a^{2} \cos ^{2} \alpha=9 a^{2}+72 a^{2} \cos ^{2} \alpha \\
D E^{2}=9 a^{2}+16 a^{2}+24 a^{2} \cos 2 \alpha=25 a^{2}+24 a^{2}\left(2 \cos ^{2} \alpha-1\right)=a^{2}+48 a^{2} \cos ^{2} \alpha
\end{gathered}
$$

Using the formula for the median from triangle $D E C$, we find that

$$
\begin{gathered}
C Q^{2}=\frac{1}{4}\left(2 C E^{2}+2 C D^{2}-D E^{2}\right)=\frac{1}{4}\left(18 a^{2}+144 a^{2} \cos ^{2} \alpha+8 a^{2}-\frac{23}{4}\right)= \\
=\frac{1}{4}\left(26 a^{2}+144 a^{2} \cos ^{2} \alpha-\frac{23}{4}\right)
\end{gathered}
$$

Given that $D E=\frac{\sqrt{23}}{2}$ and $C Q=\frac{\sqrt{23}}{2}$, we have the system of equations

$$
\left\{\begin{array}{l}
a^{2}+48 a^{2} \cos ^{2} \alpha=\frac{23}{4} \\
26 a^{2}+144 a^{2} \cos ^{2} \alpha=\frac{115}{4}
\end{array}\right.
$$

Multiplying the first equation by -3 and adding the resulting equation to the second, we find that $a^{2}=\frac{1}{2}$. Then

$$
\cos ^{2} \alpha=\frac{7}{32}, \sin ^{2} \alpha=\frac{25}{32}
$$

Let $R$ be the radius of the circumcircle of triangle $A B C$. Then

$$
R=\frac{B C}{2 \sin \alpha}=\frac{6 a}{2 \sin \alpha}=\frac{3 \sqrt{2}}{\frac{5 \sqrt{2}}{4}}=\frac{12}{5} \text {. }
$$