$$
\begin{aligned}
& {[\text { Orthogonal projection (other). ] }} \\
& \text { [ Properties and characteristics of a parallelogram ] }
\end{aligned}
$$

The distances from three consecutive vertices of a parallelogram to a certain plane are 1, 3, and 5. Find the distance from the fourth vertex to this plane.
Let $A 1, B 1, C 1, D 1, O 1$ be the orthogonal projections of the vertices $A, B, C, D$ of parallelogram $A B C D$ and its center $O$ onto plane $\alpha$, with $A A 1=1, B B 1=3, C C 1=5$. If points $A$ and $C$ lie on the same side of plane $\alpha$, then $O O 1$ is the midline of the right trapezoid $A A 1 C 1 C$, so

$$
O O 1=\frac{1}{2}(A A 1+C C 1)=\frac{1}{2}(1+5)=3
$$

If point $B$ also lies on the same side of plane $\alpha$ as points $A$ and $C$ (Fig.1), then line $B O$ is parallel to plane $\alpha (B B 1=O O 1=3)$, so point $D$, lying on this line, is at a distance of 3 from plane $\alpha$. If point $D$ lies on the same side of plane $\alpha$ as points $A$ and $C$, and point $B$ lies on the opposite side (Fig.2), then segment $O O 1$ connects the midpoints of the diagonals $B D$ and $B 1 D 1$ of trapezoid $B D 1 D B 1$ with bases $B B 1$ and $D D 1$. Therefore,

$$
O O 1=\frac{1}{2}(D D 1-B B 1), \text { or } 3=\frac{1}{2}(D D 1-3)
$$

from which we find that $D D 1=9$. If points $A$ and $C$ lie on opposite sides of plane $\alpha$, then segment $O O 1$ connects the midpoints of the diagonals $A C$ and $A 1 C 1$ of trapezoid $A A 1 C C 1$ with bases $A A 1$ and $C C 1$. Therefore,

$$
O O 1=\frac{1}{2}(C C 1-A A 1)=\frac{1}{2}(5-1)=2
$$

and point $O$ lies on the same side of plane $\alpha$ as point $C$. If point $B$ lies on the same side of plane $\alpha$ as point $A$, and point $D$ lies on the opposite side (Fig.3), then segment $O O 1$ connects the midpoints of the diagonals $B D$ and $B 1 D 1$ of trapezoid $B D 1 D B 1$ with bases $B B 1$ and $D D 1$. Therefore,

$$
O O 1=\frac{1}{2}(D D 1-B B 1), \text { or } 2=\frac{1}{2}(D D 1-3)
$$

from which we find that $D D 1=7$. If points $B$ and $D$ lie on the same side of plane $\alpha$ as point $C$ (Fig.4), then segment $O O 1$ is the midline of trapezoid $B B 1 D 1 D$ with bases $B B 1$ and $D D 1$. Therefore,

$$
O O 1=\frac{1}{2}(D D 1+B B 1), \text { or } 2=\frac{1}{2}(D D 1+3)
$$

from which we find that $D D 1=1$. The case where points $A$ and $D$ lie on the same side of plane $\alpha$, and points $B$ and $C$ lie on the opposite side, is impossible.

## Answer

$1 ; 3 ; 7 ; 9$

At vertex $A$ of rectangle $A B C D$ with sides $A B=a, B C=b$, a spider sits, and a fly is in the opposite vertex. They are separated by a vertical wall in the form of an isosceles triangle $B M D$ with base $B D$ and angle $\alpha$ at vertex $M$. Find the length of the shortest path from the spider to the fly, given that the spider can move only on the part of the plane of the rectangle where the wall is (including the boundary of the rectangle) and on the wall itself.

## Solution

Let $\angle A D B=\beta$. Suppose the spider's path consists of consecutive segments $A P, P Q, Q P$, and $P C$, where point $P$ lies on the diagonal $B D$ of rectangle $A B C D$, and point $Q$ lies on the side, for example, $B M$, of triangle $B M D$. Consider the unfolding of this configuration, consisting of a right triangle $A B D$, an isosceles triangle $B M D$, an isosceles triangle $B M D '$ equal to it, and a right triangle $B D^{\prime} C^{\prime}$ with a right angle at vertex $C^{\prime}$, equal to triangle $D A B$. In this case, any two adjacent of the four specified triangles have exactly one common side, point $P$ corresponds to points $P^{\prime}$ and $P^{\prime \prime}$ on segments $B D$ and $B D^{\prime}$, and point $Q$ corresponds to point $Q^{\prime}$ on segment $B M$. If $\alpha \geqslant 90^{\circ}$, then all points of segment $A C^{\prime}$ belong to the unfolding. Then

$$
\begin{gathered}
A P^{\prime}+P^{\prime} Q^{\prime}+Q^{\prime} P^{\prime \prime}+P^{\prime \prime} C^{\prime} \geqslant A C^{\prime} \\
\angle A B C^{\prime}=\angle A B D+\angle D B M^{\prime}+\angle M^{\prime} B D^{\prime}+\angle D^{\prime} B C^{\prime}= \\
=90^{\circ}-\beta+2\left(90^{\circ}-\frac{1}{2} \alpha\right)+\beta=270^{\circ}-\alpha \leqslant 180^{\circ} .
\end{gathered}
$$

Thus, the shortest path sought is equal to the length of segment $A C^{\prime}$. By the cosine theorem from triangle $A B C^{\prime}$, we find that

$$
\begin{gathered}
A C^{\prime}=\sqrt{A B^{2}+B C^{\prime 2}-2 A B \cdot B C^{y} \cos \angle A B C^{\prime}}= \\
=\sqrt{\left.a^{2}+b^{2}-2 a b \cos (27)^{\circ}-\alpha\right)}=\sqrt{a^{2}+b^{2}+2 a b \sin \alpha}
\end{gathered}
$$

If $\alpha<90^{\circ}$, then the shortest path will pass along sides $A B$ and $B C$ (or $A D$ and $D C$) and will be equal to $a+b$.

## Answer

$\sqrt{a^{2}+b^{2}+2 a b \sin \alpha}$, if $90^{\circ} \leqslant \alpha<180^{\circ} ; a+b$, if $\alpha<90^{\circ}$.