$\mathrm{Az} A B C D$ rectangular base pyramid $E$ has its vertex $E$ above vertex $A$. $A B=4 \mathrm{~cm}$. What is the length of edge $B C$, if $B C E \varangle=60^{\circ}$, $A D E \varangle=45^{\circ}$.
The line $E A$ is perpendicular to the plane $A B C D$, and thus to every line in that plane, including $B C$. The line $B C$ is therefore perpendicular to $A E$, and also to $A B$, which are two non-parallel lines in the plane $A B E$; hence, $B C$ is perpendicular to the plane $A B E$; consequently, $B C$ forms a right angle with $B E$.

![](https://cdn.mathpix.com/cropped/2024_05_02_d6e4ab133dde63d52bebg-1.jpg?height=509&width=681&top_left_y=304&top_left_x=711)

Denoting the equal lengths of the edges $B C$ and $A D$ by $x$, from the isosceles right triangle $A D E$ we have $A E = x$. On the other hand, from the right triangle $B C E$ with a $60^{\circ}$ angle at $C$, we have $B E = B C \sqrt{3} = x \sqrt{3}$. Thus, from the right triangle $A B E$,

$$
x^{2} + 16 = 3 x^{2}, \quad x = \sqrt{8} \approx 2.83 \text{ cm}
$$

Iren Fauszt (Budapest, Móricz Zs. g. III. o. t.)

Remarks. 1. Similarly, we can show that the triangle $C D E$ is right-angled at $D$, and we can also calculate from this. It also turns out that this triangle is isosceles.

2. Many contestants used the fact that the angle $C B E$ (or $C D E$) is a right angle without proof. Some justified the necessary perpendicularities by drawing the net of the pyramid. In this case, the correctness of the drawing must be proven (and it can be: the projections of the arcs described by the vertex $E$ onto the base are perpendicular to the edge used as the axis).

![](https://cdn.mathpix.com/cropped/2024_05_02_d6e4ab133dde63d52bebg-1.jpg?height=509&width=601&top_left_y=1426&top_left_x=689)

Figure 2