[ Area and orthogonal projection ]

The area of the orthogonal projection of a circle with a radius of 1 onto the plane $\alpha$ is 1. Find the length of the orthogonal projection of this circle onto a line perpendicular to the plane $\alpha$.

#
Let $\phi$ be the angle between the plane $\alpha$ and the plane of the given circle, $S$ be the area of the given circle with radius $r=1$, and $S_1$ be the area of its orthogonal projection onto the plane $\alpha$. Then, by the theorem of the area of the orthogonal projection,

![](https://cdn.mathpix.com/cropped/2024_05_06_970e9bfb17ad7b5a0e46g-14.jpg?height=65&width=726&top_left_y=2354&top_left_x=672)

from which we find that $\cos \phi=\frac{1}{\pi}$. Then,

$$
\sin \phi=\sqrt{1-\cos ^{2} \varphi}=\sqrt{1-\frac{1}{\pi^{2}}}=\frac{\sqrt{\pi^{2}-1}}{\pi}
$$

Let $a$ be the orthogonal projection of the given circle onto the line $l$, which is perpendicular to the plane $\alpha$. Then,

$$
a=2 r \cos \left(90^{\circ}-\phi\right)=2 r \sin \phi=\frac{2 \sqrt{\pi^{2}-1}}{\pi}
$$

Problem