30. Two Projections. Imagine a plane $\Pi_{1}$ tangent to the Earth at the North Pole $N$, and a plane $\Pi_{2}$ tangent to the Earth at the South Pole $S$. We can draw one map by projecting each point on the Earth's surface from $N$ onto $\Pi_{2}$, and another map by projecting each point from $S$ onto $\Pi_{1}$; this is known as the stereographic projection. Now we can fold both planes so that the meridians align. Each point on one map corresponds to a specific point on the other map, thus defining a certain mapping of the plane onto itself. How can this mapping be directly determined?
30. It is easy to verify that the transformation in question in the problem is an inversion *). Indeed, let $\Pi_{1}$ and $\Pi_{2}$ be the tangent planes to the Earth's sphere at points $N$ and $S$ (Fig. 39). Suppose, according to the transformation defined in the problem, point $P_{1}$ on plane $\Pi_{1}$ and point $\dot{P}_{2}$ on plane

![](https://cdn.mathpix.com/cropped/2024_05_21_5d77f4829f88503a49eag-082.jpg?height=551&width=688&top_left_y=821&top_left_x=381)

Fig. 39.

![](https://cdn.mathpix.com/cropped/2024_05_21_5d77f4829f88503a49eag-082.jpg?height=434&width=605&top_left_y=957&top_left_x=1111)

Fig. 40.

$\Pi_{2}$ correspond to each other. Consider the section of the Earth's sphere and planes $\Pi_{1}$ and $\Pi_{2}$ by a plane passing through the axis $N S$ (Fig. 40). Denoting $N P_{1}=r_{1}, S P_{2}=r_{2}, N S=2 r$, we obtain from the similarity of the right triangles $N T P_{1}, S T N, P_{2} T S$ the equality $r_{1} r_{2}=4 r^{2}$, which defines the inversion: if we "raise" plane $\Pi_{2}$, aligning it with $\Pi_{1}$, then points $P_{1}$ and $P_{2}$ will lie on the same line with the center $N \equiv S$ of the inversion and $N P_{1} \cdot N P_{2}=4 r^{2}=$ $=$ const.