In the cube $A B C D A 1 B 1 C 1 D 1$, where $A A 1, B B 1, C C 1$ and $D D 1$ are parallel edges, the plane $P$ passes through point $D$ and the midpoints of edges $A 1 D 1$ and $C 1 D 1$. Find the distance from the midpoint of edge $A A 1$ to the plane $P$, if the edge of the cube is 2.
Let $M, N, K$ and $L$ be the midpoints of edges $A1D1, C1D1$, $AA1$ and $CC1$ respectively (Fig.1). The line $KL$ is parallel to the line $MN$, so the line $KL$ is parallel to the plane $P$. Moreover, the line $KL$ passes through the center $O$ of the cube $ABCD A1B1C1D1$. Therefore, the distance from the midpoint $K$ of edge $AA1$ to the plane $P$ is equal to the distance from point $O$ to this plane. Let $E$ be the intersection point of segments $MN$ and $B1D1$, and $F$ be the foot of the perpendicular dropped from point $O$ to line $DE$. Since line $MN$ is perpendicular to the plane $BB1D1D$, then $MN \perp OF$. Therefore, $OF$ is the perpendicular to the plane $P$. Thus, the desired distance is equal to the length of segment $OF$. Consider the section of the given cube by the plane $BB1D1D$ (Fig.2). Let the line passing through the midpoint $O$ of diagonal $BD1$ parallel to $BD$ intersect segment $DE$ at point $Q$, and segment $DD1$ at point $S$. Then $OS$ is the midline of triangle $BD1D$, and $QS$ is the midline of triangle $EDD1$. Therefore,

$$
OQ = OS - QS = \frac{1}{2} BD - \frac{1}{2} ED1 = \frac{1}{2}(BD - ED1) = \frac{1}{2}\left(2\sqrt{2} - \frac{\sqrt{2}}{2}\right) = \frac{3\sqrt{2}}{1}
$$

Let $\angle FOQ = \alpha$. Then

$$
\angle EDD1 = \angle FOQ = \alpha, \cos \alpha = \frac{DD1}{DE} = \frac{DD1}{\sqrt{DD1^2 + ED1^2}} = \frac{2}{\sqrt{1 + \left(\frac{\left(\bar{V}_2^2\right.}{2}\right)^2}} = \frac{2\sqrt{2}}{3}
$$

Therefore,

$$
OF = OQ \cos \alpha = \frac{3\sqrt{2}}{1} \cdot \frac{2\sqrt{2}}{3} = 1
$$

Choose a rectangular coordinate system $D1xyz$, taking the origin at point $D1$ and directing the coordinate axes along the rays $D1A1, D1C1$, and $D1D$ respectively. The equation of the plane $P$ will be

$$
\frac{x}{1} + \frac{y}{1} + \frac{z}{2} = 1
$$

(the equation of the plane in intercept form), or $2x + 2y + z - 2 = 0$. The distance from point $K(2; 0; 1)$ to this plane is

$$
\frac{2 \cdot 2 + 2 \cdot 0 + 1 - 2}{\sqrt{1 + 1 + 1}} = \frac{3}{3} = 1
$$

## Answer

1.00