[The ratio in which the bisector divides the side] Euler's formula

In triangle $A B C$, the bisectors $A E$ and $C D$ are drawn. Find the lengths of the segments $C D, C E, D E$ and the distance between the centers of the circle inscribed in triangle $A B C$ and the circle circumscribed around triangle $A B C$, if $A C=2$

$$
, B C=4, \angle A C B=\arccos \frac{11}{16}
$$
Let $\angle A C B=\gamma, \angle A B C=\beta$. According to the problem, $\cos \gamma=\frac{11}{16}$. Then

$$
\sin \gamma=\sqrt{1-\cos ^{2} \gamma}=\sqrt{1-\frac{121}{256}}=\frac{3 \sqrt{15}}{16}.
$$

By the cosine rule

$$
A B=\sqrt{A C^{2}+B C^{2}-2 A C \cdot B C \cos \gamma}=\sqrt{4+16-2 \cdot 2 \cdot 4 \cdot \frac{11}{16}}=\sqrt{20-11}=3.
$$

By the property of the angle bisector of a triangle

$$
\begin{gathered}
\frac{A D}{D B}=\frac{A C}{B C}=\frac{2}{4}=\frac{1}{2}, \frac{C E}{B E}=\frac{A C}{A B}=\frac{2}{3}, \\
\text { therefore } \\
A D=\frac{1}{3} A B=1, B D=\frac{2}{3} A B=2, C E=\frac{2}{5} B C=\frac{2}{5} \cdot 4=\frac{8}{5}, B E=\frac{12}{5}.
\end{gathered}
$$

By the formula for the angle bisector of a triangle $C D=\sqrt{A C \cdot B C-A D \cdot B D}=\sqrt{2 \cdot 4-1 \cdot 2}=\sqrt{6}$.

Let $R$ be the radius of the circumcircle of triangle $A B C$. By the sine rule

$$
R=\frac{A B}{2 \sin \gamma}=\frac{3}{2 \cdot \frac{3 \sqrt{15}}{16}}=\frac{8}{\sqrt{15}}, \sin \beta=\frac{A C}{2 R}=\frac{2}{2 \cdot \frac{8}{\sqrt{15}}}=\frac{\sqrt{15}}{8}
$$

Then $\cos \beta=\frac{7}{8}$. From triangle $B D E$ by the cosine rule we find that

$$
D E=\sqrt{B D^{2}+B E^{2}-2 B D \cdot B E \cos 3}=\sqrt{4+\frac{144}{25}-2 \cdot 2 \cdot \frac{12}{5} \cdot \frac{7}{8}}=\frac{\sqrt{34}}{5}.
$$

Let $r$ be the radius of the incircle of triangle $A B C, p$ be the semiperimeter of the triangle, and $S$ be the area. Then

$$
\begin{gathered}
p=\frac{2+3+4}{2}=\frac{9}{2}, S=\frac{1}{2} A C \cdot B C \sin \gamma=\frac{1}{2} \cdot 2 \cdot 4 \cdot \frac{3 \sqrt{15}}{16}=\frac{3 \sqrt{15}}{4}. \\
\text { Therefore, } \\
r=\frac{S}{p}=\frac{\frac{3 \sqrt{15}}{4}}{\frac{9}{2}}=\frac{\sqrt{15}}{6}.
\end{gathered}
$$

Let the incircle with center $Q$ touch side $A C$ of triangle $A B C$ at point $M$, and $N$ be the midpoint of this side. Then

By the Pythagorean theorem

$$
O N=\sqrt{O C^{2}-C N^{2}}=\sqrt{R^{2}-C N^{2}}=\sqrt{\frac{64}{15}-1}=\frac{7}{\sqrt{15}}
$$

The desired segment $O Q$ is the longer lateral side of the right trapezoid $M N O Q$ with bases $M Q$ and $O N$.

Let $F$ be the foot of the perpendicular dropped from point $Q$ to $O N$. Then

$$
\begin{gathered}
O F=O N-F N=O N-Q M=\frac{7}{\sqrt{15}}-\frac{5}{2 \sqrt{15}}=\frac{9}{2 \sqrt{15}}, Q F=M N=\frac{1}{2}. \\
\rho=O Q=\sqrt{Q F^{2}+O F^{2}}=\sqrt{\frac{1}{4}+\frac{81}{60}}=2 \sqrt{\frac{2}{5}}.
\end{gathered}
$$

Hint. To find the distance between the centers of the inscribed and circumscribed circles of a triangle, you can use Euler's formula:

$$
\rho=\sqrt{R^{2}-2 R r}
$$

## Answer

$C D=\sqrt{6}, C E=\frac{8}{5}, D E=\frac{\sqrt{34}}{5}, \rho=2 \sqrt{\frac{2}{5}}$.