A plane figure's orthogonal projection on two perpendicular planes is a 2 unit side square on each. What is the perimeter of the plane figure if one of its sides is $\sqrt{5}$ units?
Since the orthogonal projections of the opposite sides of the quadrilateral on the two intersecting planes are parallel in each plane, the opposite sides of the quadrilateral are also parallel, so the quadrilateral is a parallelogram. Move the parallelogram so that the vertex which was closest to the intersection line of the two planes coincides with the intersection line. The projections of the translated parallelogram on the planes are congruent to the projections of the original parallelogram.

![](https://cdn.mathpix.com/cropped/2024_05_02_c09113717ec97226ea4dg-1.jpg?height=520&width=594&top_left_y=350&top_left_x=749)

Figure 1

Let the vertex of the parallelogram on the intersection line of the two planes be denoted by $A$, the vertex at a distance of $\sqrt{5}$ from $A$ by $B$, the other vertex adjacent to $A$ by $D$, and the vertex opposite to $A$ by $C$. Let the orthogonal projections of the points $B, C, D$ on one of the planes be $B^{\prime}, C^{\prime}, D^{\prime}$; and the orthogonal projections of $B^{\prime}$ and $D^{\prime}$ on the intersection line of the two planes be $B^{\prime \prime}$ and $D^{\prime \prime}$. Finally, let $A D=\alpha$.

According to our conditions, $A B=\sqrt{5}$ and $A B^{\prime}=2$, so $B B^{\prime}=\sqrt{A B^{2}-A B^{\prime 2}}=1$, which means $B$ is also 1 unit away from the other plane, since the projection of the segment $A B$ on the other plane is also 2 units long; therefore, $B^{\prime} B^{\prime \prime}=1$.

The triangles $A B^{\prime} B^{\prime \prime}$ and $A D^{\prime} D^{\prime \prime}$ are congruent because $A B^{\prime}=A D^{\prime}=2$, $\angle A B^{\prime \prime} B^{\prime}=\angle A D^{\prime \prime} D^{\prime}=90^{\circ}$, and $\angle B^{\prime} A B^{\prime \prime}=\angle D^{\prime \prime} D^{\prime} A$ are right-angled angles. Thus, $D^{\prime} D^{\prime \prime}=B^{\prime \prime} A=\sqrt{A B^{\prime 2}-B^{\prime} B^{\prime \prime 2}}=\sqrt{3}$. On the other hand, the point $D$ is also at the same distance from both planes, so $D^{\prime} D^{\prime \prime}=D^{\prime} D=\sqrt{A D^{2}-A D^{\prime 2}}=\sqrt{a^{2}-4}$. Therefore, $\sqrt{3}=\sqrt{a^{2}-4}$, from which $a=\sqrt{7}$.

The perimeter of the parallelogram is thus $2(A B+A D)=2(\sqrt{5}+\sqrt{7})$.

Based on the work of Krisztián Németh (Bp., Fazekas M. Gyak.Gimn., II. o. t.).

Remarks: 1. We only used the fact that the projection of the triangle $A B D$ on the two planes is an isosceles right triangle. In this view, the quadrilateral is in one of the angle bisector planes of the two planes. However, the desired vertex $A$ does not exist for every possible position of the quadrilateral and the planes. Specifically, if the plane of the quadrilateral is parallel to the other angle bisector plane of the two planes. In this case, the sum of the distances from the points $A, B, C, D$ to the two planes are equal (Figure 2). In this case, it is advisable to move one of the planes so that it passes through $A$ (or $C$).

![](https://cdn.mathpix.com/cropped/2024_05_02_c09113717ec97226ea4dg-1.jpg?height=558&width=634&top_left_y=1792&top_left_x=732)

2. From our point of view, the order of the vertices of the projections $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ and $A^{*} B^{*} C^{*} D^{*}$ are in opposite directions on Figure 1, but they are in the same direction on Figure 2.
3. It can be shown by simple calculation that the diagonals $A C$ and $B D$, which appear to be of equal length from both directions, are not equal, $A C > B D$.