Given a cube $A B C D A 1 B 1 C 1 D 1$ with edge $a$. Let $M$ be a point on the edge $A 1 D 1$ such that $A 1 M: M D 1=1: 2$. Find the perimeter of triangle $A B 1 M$, as well as the distance from vertex $A 1$ to the plane passing through the vertices of this triangle.
From the right triangles $A A_1 M, B_1 A_1 M$, and $A A_1 B_1$, using the Pythagorean theorem, we find that

\[
\begin{gathered}
A M = \sqrt{A A_1^2 + A_1 M^2} = \sqrt{\boldsymbol{a}^2 + \frac{\alpha^2}{9}} = \frac{a \sqrt{10}}{3} \\
B_1 M = \sqrt{A_1 B_1^2 + A_1 M^2} = \sqrt{\boldsymbol{a}^2 + \frac{\alpha^2}{9}} = \frac{a \sqrt{10}}{3} \\
A B_1 = \sqrt{A_1 A^2 + A_1 B_1^2} = \sqrt{\boldsymbol{a}^2 + \sigma^2} = a \sqrt{2}
\end{gathered}
\]

Therefore, the perimeter of triangle $A B_1 M$ is

\[
\begin{gathered}
A M + B_1 M + A B_1 = \frac{a \sqrt{10}}{3} + \frac{a \sqrt{10}}{3} + a \sqrt{2} = \\
= \frac{22 \sqrt{10}}{3} + a \sqrt{2} = \frac{a (2 \sqrt{10} + 3 \sqrt{2})}{3}
\end{gathered}
\]

Now let's find the distance from vertex $A_1$ to the plane passing through the vertices of triangle $A B_1 M$.

![](https://cdn.mathpix.com/cropped/2024_05_06_00e2c6ab04990c8ec18fg-28.jpg?height=46&width=263&top_left_y=1183&top_left_x=14)

Let $K$ be the midpoint of the base $A B_1$ of the isosceles triangle $M A B_1$. Then $M K$ is the height of triangle $M A B_1$. Using the Pythagorean theorem in the right triangle $M A K$, we find that

\[
M K = \sqrt{A M^2 - A K^2} = \sqrt{\frac{19 a^2}{9} - \frac{\alpha^2}{2}} = \frac{a \sqrt{11}}{3 \sqrt{2}}
\]

Let $V$ be the volume of the pyramid $A A_1 M B_1$. Then

\[
V = \frac{1}{3} S_{\triangle A_1 M B_1} \cdot A A_1 = \frac{1}{3} \cdot \frac{1}{2} A_1 M \cdot A_1 B_1 \cdot A A_1 = \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{a}{3} \cdot a \cdot a = \frac{a^3}{18}
\]

On the other hand,

\[
V = \frac{1}{3} S_{\triangle A B_1 M} \cdot h
\]

where $h$ is the desired distance from vertex $A_1$ to the plane passing through the vertices of triangle $A B_1 M$. Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_00e2c6ab04990c8ec18fg-28.jpg?height=119&width=658&top_left_y=1850&top_left_x=700)

![](https://cdn.mathpix.com/cropped/2024_05_06_00e2c6ab04990c8ec18fg-28.jpg?height=51&width=263&top_left_y=2041&top_left_x=17)

Let's take the origin at vertex $A_1$, and direct the coordinate axes along the rays $A_1 D_1, A_1 B_1$, and $A_1 A$. Then the equation of the plane passing through points $A, B_1$, and $M$ is

\[
\frac{\pi}{\frac{\pi}{x}} + \frac{y}{a} + \frac{z}{a} = 1
\]

(the equation of the plane in intercept form), or $3 x + y + z - a = 0$. Therefore,

\[
h = \frac{|a + 0 + 0 - a|}{\sqrt{9 + 1 + 1}} = \frac{a}{\sqrt{11}}.
\]

## Answer

\[
\frac{a (2 \sqrt{10} + 3 \sqrt{2})}{3} ; \frac{a}{\sqrt{11}}
\]