\section*{Problem 2 - 061042}

Given the degree measure of the angle of inclination between two planes \(\epsilon\) and \(\epsilon_{1}\). Also given is the area \(I_{\triangle A B C}\) of a triangle \(\triangle A B C\) that lies in the plane \(\epsilon\).

The feet of the perpendiculars from \(A, B, C\) to \(\epsilon_{1}\) form a (possibly degenerate) triangle \(\Delta A_{1} B_{1} C_{1}\).

What is the area \(I_{\Delta A_{1} B_{1} C_{1}}\) of this triangle?
Let \(\phi\) be the angle between the two planes and \(g\) their line of intersection.

By orthogonal projection, segments in \(\epsilon\), which are parallel to \(g\), are mapped to congruent segments in \(\epsilon_{1}\).

For segments in \(\epsilon\) that are perpendicular to \(g\), their images in \(\epsilon_{1}\) are stretched by a factor of \(\cos \phi\) (or, since \(\cos \phi \leq 1\), rather "compressed"). The images of lines parallel to \(g\) in \(\epsilon\) remain parallel to \(g\) in \(\epsilon_{1}\); similarly, lines orthogonal to \(g\) in \(\epsilon\) are also mapped to lines orthogonal to \(g\) in \(\epsilon_{1}\).

We consider the triangle \(\triangle A B C\) in \(\epsilon\). If one of the sides of the triangle (w.l.o.g. \(A B\)) is parallel to \(g\), then its image (i.e., \(A_{1} B_{1}\)) is congruent to it, while its height (\(h_{A B}\)) perpendicular to \(g\) has an image (\(h_{A_{1} B_{1}}\)) that is stretched by a factor of \(\cos \phi\). Since the images of these two segments are also perpendicular to each other in the target plane, the area of the image triangle is calculated as

\[
I_{\triangle A_{1} B_{1} C_{1}}=\frac{1}{2} \cdot\left|A_{1} B_{1}\right| \cdot\left|h_{A_{1} B_{1}}\right|=\frac{1}{2} \cdot|A B| \cdot \cos \phi \cdot h_{A B}=\cos \phi \cdot I_{\triangle A B C}
\]

If, on the other hand, no side of the triangle \(\triangle A B C\) is parallel to \(g\), then exactly one of the parallels in \(\epsilon\) to \(g\) through the three vertices intersects the opposite side of the triangle at an interior point.

Let this be the parallel through \(A\); and let the intersection point with \(B C\) be \(D\). Then \(A D\) is parallel to \(g\) and the heights from \(B\) and \(C\) to \(A D\) are perpendicular to \(g\), so the above considerations can be applied to each of these sub-triangles, i.e., (with \(D_{1}\) as the image point of \(D\) with respect to the orthogonal projection)

\[
I_{\triangle A_{1} B_{1} D_{1}}=\cos \phi \cdot I_{\triangle A B D} \quad \text { and } \quad I_{\triangle A_{1} C_{1} D_{1}}=\cos \phi \cdot I_{\triangle A C D}
\]

Finally, it is to be noted that the two triangles \(\triangle A B C\) and \(\triangle A_{1} B_{1} C_{1}\) are composed of the respective two sub-triangles \(\triangle A B D\) and \(\triangle A C D\) or, correspondingly, \(\triangle A_{1} B_{1} D_{1}\) and \(\triangle A_{1} C_{1} D_{1}\), so their areas are the sum of those of the respective two sub-triangles, and thus in any case:

\[
I_{\triangle A_{1} B_{1} C_{1}}=\cos \phi \cdot I_{\triangle A B C}
\]