[ Construction on a projection drawing ]

The base of a right prism $A B C A 1 B 1 C 1$ is a right triangle $A B C\left(\angle B=90^{\circ}, A B=B C=10\right)$; $A A 1=B B 1=C C 1=12$. Point $M$ is the midpoint of the lateral edge $A A 1$. A plane is drawn through points $M$ and $B 1$, forming an angle of $45^{\circ}$ with the base plane and intersecting the edge $C C 1$ at point $E$. Find $C E$.
Let the lines $M E$ and $A 1 C 1$ intersect at point $P$. Then the plane of the section intersects the plane of the base $A 1 B 1 C 1$ along the line $B 1 P$ (Fig.1). Suppose that point $P$ lies on the extension of the edge $A 1 C 1$ beyond point $A 1$. Drop the perpendicular $A 1 K$ from point $A 1$ to the line $B 1 P$. By the theorem of three perpendiculars, $M K \perp B 1 P$, so $A 1 K M$ is the linear angle of the dihedral angle between the cutting plane and the base plane of the prism. According to the problem, $\angle A 1 K M=45^{\circ}$. From the right triangles $A 1 B 1 C 1$, $A 1 K M$, and $A 1 K B 1$, we find that

$$
A 1 C 1=10 \sqrt{2}, A 1 K=A 1 M=6, B 1 K=\sqrt{A_{1} B_{1}^{2}-A_{1} K^{2}}=\sqrt{100-36}=8
$$

Consider the triangle $P B 1 C 1$ (Fig.2). Let $\angle K A 1 B 1=\alpha$. Then

$$
\cos \alpha=\frac{A_{1} K}{A_{1} B_{1}}=\frac{6}{10}=\frac{3}{5}, \sin \alpha=\frac{4}{5}
$$

$$
\begin{gathered}
\angle P A 1 K=180^{\circ}-45^{\circ}-\alpha=135^{\circ}-\alpha \\
P A 1=\frac{K A_{1}}{\cos \angle P^{P} A_{1} K}=\frac{6}{\operatorname{css}\left(135^{\circ}-\operatorname{co}\right)}=\frac{6}{\cos 135^{\circ} \cos \left(2+\sin 135^{\circ} \sin \boldsymbol{r}\right.}= \\
\frac{6 \sqrt{2}}{-\sqrt{2}+\frac{1}{5}}=30 \sqrt{2}, P C 1=P A 1+A 1 C 1=30 \sqrt{2}+10 \sqrt{2}=40 \sqrt{2}
\end{gathered}
$$

From the similarity of triangles $P E C 1$ and $P M A 1$, we find that

$$
E C 1=M A 1 \cdot \frac{P C_{1}}{P A_{1}}=6 \cdot \frac{40 \sqrt{2}}{30 \sqrt{2}}=6 \cdot \frac{4}{3}=8
$$

Therefore,

$$
C E=C C 1-E C 1=12-8=4 .
$$

If point $P$ lies on the extension of the edge $A 1 C 1$ beyond point $C 1$, then, reasoning similarly, we get that

$$
\begin{aligned}
& \cos \alpha=\frac{4}{5}, \sin \alpha=\frac{3}{5}, \\
& \angle P A 1 K=45^{\circ}+90^{\circ}-\alpha=135^{\circ}-\alpha, \\
& P A 1=\frac{K A_{1}}{\cos \angle P^{\prime} A_{1} K}=\frac{6}{\cos \left(135^{\circ}-\operatorname{r}\right)}=\frac{6}{\cos 135^{\circ} \cos r 2+\sin 135^{\circ} \sin \boldsymbol{n} \boldsymbol{n}}=
\end{aligned}
$$

which is impossible.

## Answer

4.00

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