তল ও গোলকের ছেদ

Given a plane and a sphere, drop a perpendicular from the center of the sphere (call this O) to the plane.

Call the length of this segment d, denote the foot of the altitude P, and call the radius of the sphere R.

We know that d is the minimal distance from O to any point in the plane.

We recognize the following 3 cases:

1. d>R

Since d>R, the point P lies outside the sphere.

However, since d is the minimal distance to the plane, every point X on the plane has a distance from O greater than or equal to d. Thus, OX>R for all points X, which means that all points on the plane lie outside the sphere.

Thus, the sphere and plane have no intersection.

1. d=R

Now, the point P lies on the sphere.

By a similar reasoning to case 1, every point X on the plane (other than P) satisfies OX>R. Thus, every other point lies outside the sphere, and the sphere and the plane have exactly one point of intersection: the point P.

In this case, the sphere is tangent to the plane.

1. d<R

This is the most interesting case. Consider a circle within the plane, with center P and radius r such that $r^2=R^2-d^2$.

By the Pythagorean Theorem, a point A on that circle will lie a distance R from O. Thus, all points on that circle also lie on the sphere.

This means that the sphere and plane intersect in a circle with center P and radius r such that $r^2=R^2-d^2$.

A great circle of a sphere is the intersection of the sphere and a plane that passes through the center of the sphere. Great circles have several noteworthy properties.

Great circles are the spherical geometry equivalent of lines in Euclidean geometry. Any two distinct great circles meet in exactly 2 diametrically opposite points. The fact that no two lines are parallel leads to the modified version of Playfair's Axiom, and it is this postulate that distinguishes elliptic geometry.

For any two points on a sphere, the shortest path between the two points is an arc of a great circle (this path is called a geodesic).

To prove this, place this sphere in 3-dimensional spherical coordinate space.

Call the two points A and B, and let the radius of the sphere be R.

Position the sphere such that A lies at the "North Pole" (R,0,0) (although technically, $\theta$ is arbitrary at this point).

Rotate the sphere such that B lies at a point (R,0,$\phi$).

The distance from A to B along the great circle is R$\phi$. Along this arc, the value of $\theta$ is always zero.

Now suppose that we travel from A to B along some other arc. Then, $\theta$ will not be constant on the path from A to B. Thus, we must travel the same distance R$\phi$ plus an extra amount that results from the changes in $\theta$.

Thus, the shortest path lies along a great circle.

A more formal proof requires a parametrization of $\theta$ and $\phi$ on the path from A to B. One would then use calculus to compute the arc length from A to B. However, the same result is achieved: that the distance is minimized when d$\theta$/dt=0.