Let's create a paper model of the shape, choosing the side of the pentagon to be of unit length. $A B C$ and $A E D$ are isosceles right triangles, so $A C = A D = \sqrt{2}$, making $A C D$ an isosceles triangle. Given that the base $C D = 1$, the angles at $C$ and $D$ are greater than $60^{\circ}$. By constructing the triangles $A B^{*} C$, $A C D$, and $A D E^{*}$ in sequence (Figure 1), the sides of the (cut-out) pentagon $A B^{*} C D E^{*}$ are equal, and the angles at $C$ and $D$ are greater than the prescribed right angle ( $45^{\circ} + 60^{\circ}$ ).

![](https://cdn.mathpix.com/cropped/2024_05_02_479ff629c820a23c24a1g-1.jpg?height=881&width=683&top_left_y=367&top_left_x=707)

Figure 1

Fix the plane $A C D = S_{1}$, and rotate the triangle $A B^{*} C$ around $A C$, and the triangle $A D E^{*}$ around $A D$ until $C B^{*}$ is perpendicular to $C D$, and $D E$ is perpendicular to $D C$, and follow the path of the projections of $B^{*}$ and $E^{*}$ onto $S_{1}$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_479ff629c820a23c24a1g-1.jpg?height=730&width=1027&top_left_y=1470&top_left_x=533)

Figure 2

During the rotation, the projection moves along the perpendicular bisector of the segments $A C$ and $A D$. In the desired positions of $B^{\prime}$ and $E^{\prime}$, $B^{\prime} C$ and $E^{\prime} D$ are perpendicular to $C D$. From this, $B^{\prime}$ and $E^{\prime}$ can be constructed in $S_{1}$, clearly being mirror images of each other with respect to the perpendicular bisector of the segment $C D$ at $A F$, and the equality of $B^{\prime} C$ and $E^{\prime} D$ implies that the distances of $B$ and $E$ from $S_{1}$ are equal.

If both rotations would place $B$ and $E$ above $S_{1}$, or if both would place them below $S_{1}$, then $B E$ would be parallel to $C D$ and these four points would lie in the same plane. This is excluded by the problem, so $B$ and $E$ end up on opposite sides of $S_{1}$. Denoting the midpoint of the side $C D$ by $F$, the figure is mapped onto itself by a half-turn around the axis $A F$. Thus, the quadrilateral $A B F E$ is a planar kite, $E A B \angle = 2 E A F \angle$, and in the triangle $E A F$, $A F = \sqrt{7} / 2$, $E F = \sqrt{5} / 2$, so

$$
\cos E A F \angle = \frac{A E^{2} + A F^{2} - E F^{2}}{2 A E \cdot A F} = \frac{3}{2 \sqrt{7}},
$$

thus $\cos E A B \angle = -5 / 14$, and the sought angle is $110^{\circ} 55^{\prime}$.

Ernő Kulcsár (Szombathely, Nagy Lajos Gymnasium, 3rd grade)

Remark. The 5 sides and 4 angles uniquely determined our spatial (skew) pentagon. This is generally the case. By cutting a planar polygon into triangles from one of its vertices using diagonals, the first triangle is determined by 3 data, and the subsequent ones by 2 new data each, considering the common side. In space, for any such broken line, the inclination of the planes of the partial triangles to each other must somehow be given, so 3 data are needed for each new and subsequent partial triangle.
The equilateral pentagon $ABCDE$ does not have four vertices lying in the same plane. The angles $\angle ABC, \angle BCD, \angle CDE$, and $\angle DEA$ are right angles. What is the measure of the angle $\angle EAB$? (The figure appeared in the 1966 Kürschák competition, but no such calculation was involved there.)