Theorem. A plane perpendicular to a radius of a sphere at its extremity is tangent to the sphere.

Data: Any sphere, and a plane, as MN, perpendicular to a radius, as OP, at its extremity P.

To prove

MN tangent to the sphere.

Proof. Take any point except P in MN, as 4, and draw OA.

point 4 is without the sphere.

But A is any point in MN except P;

every point in MN except P is without the sphere.

MN is tangent to the sphere at P.

Q.E.D.

Hence, § 638, 670. Cor. I. Any straight line perpendicular to a radius of a sphere at its extremity is tangent to the sphere.

671. Cor. II. Any plane or line tangent to a sphere is perpendicular to the radius drawn to the point of contact.

672. Cor. III. A straight line tangent to any circle of a sphere lies in the plane tangent to the sphere at the point of contact.

673. Cor. IV. Any straight line drawn in a tangent plane and through the point of contact is tangent to the sphere at that point.

674. Cor. V. Any two straight lines tangent to a sphere at the same point determine the tangent plane at that point.

Proposition V

675. Select any four points not in the same plane; form a tetrahedron of which these points are the vertices; and at the centers of the circles circumscribed about any two of its faces erect perpendiculars. How do the distances from any point in either perpendicular to the vertices of the face to which it is perpendicular compare? If these per

pendiculars intersect, how, then, do the distances from their intersection to the four given points compare? Is there any other point that is equidistant from the four given points? What surface, then, may be passed through these four points? How many such surfaces may be passed through them?

Theorem. Through any four points not in the same plane one spherical surface may be passed, and only one. Data: Any four points not in the same plane, as A, B, C, D.

may

To prove that one spherical surface be passed through A, B, C, D, and only one.

Proof. Suppose H and G to be the centers of circles circumscribed about the triangles BCD and ACD, respectively.

Draw HK plane BCD, and GE plane ACD.

B

H

K

A

§ 450, every point in HK is equidistant from points B, C, D, and

every point in GE is equidistant from points A, C, D.

From H and G draw lines to L, the middle point of CD.

.. § 444, the plane through HL and GL is perpendicular to CD, and, § 483, this plane is perpendicular to planes BCD and ACD. Const., GEL plane ACD at G;

.. § 481,

GE lies in the plane HLG.

In like manner it may be shown that HK lies in plane HLG. Hence, the perpendiculars HK and GE lie in the same plane, and, being perpendicular to planes which are not parallel, they must intersect at some point, as at 0.

Since o is in the perpendiculars HK and GE, it is equidistant from B, C, D, and from A, C, D.

Hence, o is equidistant from A, B, C, D, and the surface of the sphere, whose center is 0 and radius OB, will pass through the points A, B, C, D.

Now, § 450, the center of any sphere whose surface passes through the four points A, B, C, D must be in the perpendiculars HK and GE.

Hence, 0, the intersection of HK and GE, must be the center of the only sphere whose surface can pass through A, B, C, D. Therefore, etc.

Q.E.D.

676. Cor. I.

hedron.

677. Cor. II.

A sphere may be circumscribed about any tetra

The four perpendiculars to the faces of a tetrahe

dron through their centers meet at the same point.

Proposition VI

678. Form any tetrahedron and pass planes bisecting any three of its dihedral angles which have one face in common. How do the distances from the point of intersection of these bisecting planes to the faces of the tetrahedron compare? What figure, then, may be inscribed in any tetrahedron ?

Theorem. A sphere may be inscribed in any tetrahedron.

To prove that a sphere may be inscribed in D-ABC.

Proof. Bisect any three of the dihedral angles which have one face common, as AB, BC, AC, by the planes OAB, OBC, OAC, respectively.

By § 488, every point in the plane OAB is equidistant from the faces ABC and ABD.

Also every point in plane OBC is equidistant from the faces ABC and BCD; and every point in the plane OAC is equidistant from the faces ABC and ACD.

Therefore, point 0, the intersection of these three planes, is equidistant from the four faces of the tetrahedron.

Hence, § 638, a sphere with O as a center and with a radius equal to the distance from 0 to any face will be tangent to each face, and, § 640, it will be inscribed in the tetrahedron.

Therefore, etc.

Q.E.D.

679. Cor. The six planes which bisect the six dihedral angles of a tetrahedron intersect in the same point.

Proposition VII

680. Problem. To find the radius of a material sphere.

Datum: Any sphere, as APP'.

Required to find the radius of APP'.

Solution. From any point P as a pole describe any circumference on the surface. § 667.

From any three points in this circumference, as A, B, C, measure the chord distances AB, BC, AC.

Construct the AA'B'C' having its sides equal respectively to AB, BC, AC, and circumscribe about it a circle.

With the radius OB' as a side, and PB' equal to the chord of the are joining P and B, as hypotenuse, construct the rt. A PB'O.

Draw a line from B' perpendicular to B'P, and produce it to meet PO produced in P'.

Then, PP' thus determined is equal to the diameter of the sphere, and its half, PG, is the required radius.

Proof. By the student.

Q.E.F.

Ex. 870. Equal straight lines whose extremities are in the surface of a sphere are equally distant from the center of the sphere.

Ex. 871. The six planes which bisect at right angles the six edges of a tetrahedron all intersect at the same point.

MILNE'S GEOM. - 22

SPHERICAL ANGLES AND POLYGONS

681. The angle between two intersecting curves is the angle contained by the two tangents to the curves at their intersection.

682. The angle between two intersecting arcs of great circles is called a Spherical Angle.

683. A portion of the surface of a sphere bounded by three or more arcs of great circles is called a Spherical Polygon.

The bounding arcs are the sides of the polygon; the angles which they form are the angles of the polygon; and the points of intersection are the vertices of the polygon.

An arc of a great circle joining any two

non-adjacent vertices of a spherical polygon is a diagonal. 684. The planes of the sides of a spherical polygon form a polyhedral angle whose vertex is the center of the sphere, and whose face angles are measured by the sides of the polygon.

O-ABCDE is a polyhedral angle whose vertex O is the center of the sphere and whose face angles EOD, DOC, etc., are measured by the sides ED, DC, etc., of the spherical polygon ABCDE.

E

B

685. A spherical polygon whose corresponding polyhedral angle is convex is called a Convex Spherical Polygon.

Spherical polygons will be regarded as convex unless otherwise specified.

686. A spherical polygon of three sides is called a Spherical Triangle.

A spherical triangle is right, oblique, equilateral, isosceles, etc., under the same conditions that plane triangles are right, oblique, etc. 687. The sides of a spherical polygon, being arcs, are usually measured in degrees, minutes, and seconds.

688. Any two points on the surface of a sphere may be joined by two arcs of a great circle, one of which will usually be greater and the other usually less than a semicircumference.

Unless otherwise stated the less arc is always meant.