VI OM - III - Task 6

Through points $ A $ and $ B $, two oblique lines $ m $ and $ n $ perpendicular to the line $ AB $ have been drawn. On line $ m $, a point $ C $ (different from $ A $) has been chosen, and on line $ n $, a point $ D $ (different from $ B $). Given the lengths of segments $ AB = d $ and $ CD = l $, and the angle $ \varphi $ formed by the oblique lines $ m $ and $ n $, calculate the radius of the sphere passing through points $ A $, $ B $, $ C $, $ D $.
Fig. 20 shows the considered figure in an orthogonal projection onto the plane determined by the lines $AB$ and $n$, hence on this drawing $AB \bot n$. For simplicity, we denote the projections of points and lines with the same letters as the points and lines themselves. Let us draw through point $A$ a line $p$ parallel to line $m$, and through point $C$ a line parallel to line $AB$; it will intersect line $p$ at point $E$. The sphere $\sigma$ denotes the surface of the sphere passing through points $ABCD$; since points $A$, $B$, $C$, $D$ do not lie in the same plane (lines $m$ and $n$ are skew), such a sphere exists and is unique. Quadrilateral $ABCE$ is a rectangle because $AB \bot m$, $CE \bot m$, and $AB \bot p$; since three vertices $A$, $B$, $C$ of this rectangle lie on the sphere $\sigma$, the fourth vertex $E$ also lies on $\sigma$. The sphere $\sigma$ intersects the plane of lines $n$ and $p$ according to a circle passing through points $A$, $D$, $E$; the center $S$ of this circle is thus the center of the circumcircle of triangle $ADE$; we will explain shortly how point $S$ was constructed on Fig. 20. The center $O$ of the sphere $\sigma$ lies on the line perpendicular to the plane $ADE$ at point $S$, and thus on a line parallel to $AB$ (since $AB \bot n$ and $AB \bot p$). On the other hand, point $O$, being equidistant from points $A$ and $B$, lies on the perpendicular bisector plane of segment $AB$, so its distance $OS$ from the plane $ADE$ is equal to $\frac{1}{2} AB$. We find point $O$ by drawing a line parallel to $AB$ through point $S$ and measuring $SO = \frac{1}{2} d$ in the direction from $A$ to $B$.

Consider triangle $ASO$; the angle at vertex $S$ is a right angle, the hypotenuse $OA$ equals the sought radius $r$ of the sphere $\sigma$, $OS = \frac{1}{2} d$, and $SA$ equals the radius of the circumcircle of triangle $ADE$, so $SA = \frac{DE}{2 \sin \measuredangle DAE}$. But from the right triangle $DEC$ ($CE$ is perpendicular to the plane $ADE$) $DE^2 = DC^2 - EC^2 = l^2 - d^2$, and angle $DAE$ equals the angle $\varphi$ between the skew lines $m$ and $n$, so $SA = \frac{\sqrt{l^2 - d^2}}{2 \sin \varphi}$.

Since in triangle $ASO$ we have $OA^2 = SO^2 + SA^2$,

and finally

The construction of point $S$ is indicated on Fig. 20 with dashed lines, and it proceeds as follows. We perform a lay of triangle $ADE$ onto the plane of the drawing, i.e., onto the plane $ABD$, meaning we rotate the plane $ADE$ around line $n$ so that point $E$ ends up at some point $(E)$ in the plane of the drawing. During such a rotation, the orthogonal projection of each point in the plane $ADE$ moves along a line perpendicular to $n$. Point $(E)$ is found at the intersection of the perpendicular dropped from point $E$ to line $n$ and the ray drawn from point $A$ forming an angle $\varphi$ (in its natural size) with the ray $AD$.

Let $(S)$ be the center of the circumcircle of triangle $AD(E)$ and let the line $A(S)$ intersect the line $D(E)$ at point $(M)$. Drawing a perpendicular from point $(M)$ to line $n$ to intersect at point $M$ with line $DE$, and then drawing a perpendicular from point $(S)$ to line $n$, we find the sought point $S$ at the intersection of this perpendicular with line $AM$.

Note that when points $A$ and $B$ and lines $m$ and $n$ are given, there are infinitely many segments $CD$ of a given length $l$ with endpoints lying on lines $m$ and $n$. There are therefore infinitely many spheres passing through $A$ and $B$ and intersecting lines $m$ and $n$ at points $C$ and $D$ such that $CD = l$. The result of our problem shows that the radii of all these spheres are equal.