Proposition 3. Theorem. 679. The shortest distance on the surface of a sphere, between any two points on the surface, is the arc of a great circle, not greater than a semi-circumference, joining the two points. Hyp. Let AB be an arc of a great O, not greater than a semi Oce, joining any two pts. A and B on the sphere; and let ACEB be any other line in the surface joining A and B. To prove arc AB <ACEB. Proof. Take any pt. D in ACEB, pass arcs of great Os through A, D and B, D; and join O, the centre of the sphere, with A, D, B. Then, since the s AOB, AOD, DOB are the faces of the triedral whose vertex is at 0, .. ZAOD + DOB > ZAOB. (565) But the s AOD, DOB, AOB are measured by the arcs AD, DB, AB, respectively. .. arc AD arc DB > arc AB. (236) Similarly, joining any pt. in ACD with A and D by arcs of great Os, the sum of the arcs is greater than arc AD; and joining any pt. in DEB with D and B by arcs of great Os, the sum is greater than arc DB. If this process is indefinitely repeated, the line from A to B, on the arcs of the great Os, will continually increase and approach the line ACEB; that is, the sum of the arcs of the great Os will approach ACEB as the limit, and will always be greater than AB. .. arc AB < ACEB. Q.E.D. Proposition 4. Theorem. 680. A plane perpendicular to a radius of a sphere at its extremity is tangent to the sphere. Hyp. Let the plane MN be to the radius OP at its extremity P. To prove MN tangent to the sphere. Proof. Take any other pt. H in the plane, and join OH. Because the is the shortest distance from a point to a plane, ... OP < OH. M .. the pt. H is without the sphere. But H is any pt. of MN except P. N (497) ... every pt. of MN except P is without the sphere. ... plane MN is tangent to the sphere at the pt. P. (659) Q.E.D. 681. COR. 1. Every straight line perpendicular to a radius at its extremity is tangent to the sphere. (659) 682. COR. 2. Every plane or line tangent to a sphere is perpendicular to the radius drawn to the point of contact. 683. Cor. 3. A straight line tangent to any circle of a sphere lies in the plane tangent to the sphere at the point of contact. 684. SCH. 1. Any straight line drawn in a tangent plane through the point of contact is tangent to the sphere at that point. 685. SCH. 2. Any two straight lines, tangent to the sphere at the same point, determine the tangent plane at that point. Proposition 5. Theorem.* 686. Through any four points not in the same plane, one spherical surface, and only one, may pass. Hyp. Let A, B, C, D be the four pts. not in the same plane. To prove that one spherical surface, and no more, may pass through A, B, C, D. Proof. Let E, H be the centres of the Os circumscribed about the As BCD, ACD, respectively. Draw EK to plane BCD, and HL | to plane ACD. B E A I Every pt. in EK is equally distant from the pts. B, C, D, and every pt. in HL is equally distant from the pts. A, C, D. Join E and H to F, the middle pt. of CD. EF and HF are each to CD. (496) (203) ... the plane through EF and HF is to CD, (500) and... this plane is to both planes BCD, ACD. (540) Since HL is to the plane ACD at H, ... HL lies in the plane EFH. (Cons.) (538) Similarly, it may be shown that EK lies in this plane. ... the Is EK, HL lie in the same plane; and, being L to planes which are not ||, cannot be ||, and .·. must meet at some pt. 0. Since O is in the Is EK and HL, it is equally distant from B, C, D, and from A, C, D. (496) ... O is equally distant from A, B, C, D; and the sphere described with O as a centre and OA as a radius, will pass through the pts. A, B, C, D. Also, since the centre of any sphere passing through the four pts. A, B, C, D, must be in the Is EK, HL, (498) ... the intersection O is the centre of the only sphere that can pass through the four given pts. Q.E.D. 687. Cor. 1. A sphere may be circumscribed about any tetraedron. 688. COR. 2. The four perpendiculars to the faces of a tetraedron through their centres meet at the same point. 689. COR. 3. The six planes which bisect at right angles the six edges of a tetraedron all intersect in the same point. Proposition 6. Theorem.* 690. A sphere may be inscribed in a given tetraedron. Hyp. Let ABCD be the given tetra edron. To prove that a sphere may be inscribed in ABCD. Proof. Bisect any three of the diedrals which have one face common, в as BC, CD, BD, by the planes OBC, OCD, OBD, respectively. Since O is in the bisector of the diedral equally distant from the faces ABC and BCD. A D BC, it is (549) In the same way O is equally distant from the faces ACD and BCD, and from BAD and BCD. ... the pt. O is equally distant from the four faces of the tetraedron. ... a sphere described with O as a centre, and with a radius equal to the common distance of O from any face, will be tangent to each face, and will be inscribed in the tetraedron. (659) Q.E. D. 691. COR. The six planes which bisect the six diedral angles of a tetracdron intersect in a point. SPHERICAL TRIANGLES AND POLYGONS. DEFINITIONS. 692. The angle between two intersecting curves is the angle included between their tangents at the point of intersection. When the two curves are arcs of great circles the angle is called a spherical angle. 693. A spherical polygon is a portion of the surface of a sphere bounded by three or more arcs of great circles. The bounding arcs are the sides of the polygon; the points of intersection of the sides are the vertices of the polygon; and the angles which the sides make with each other are the angles of the polygon. A diagonal of a spherical polygon is an arc of a great circle joining any two vertices which are not consecutive. 694. A spherical triangle is a spherical polygon of three sides. A spherical triangle is right or oblique, scalene, isosceles, or equilateral, in the same cases as a plane triangle. NOTE.-Between any two points two arcs of great circles may be drawn, the one less, and the other greater, than a semi-circumference. In the present treatise the arcs less than a semi-circumference will be taken, unless otherwise stated. 695. A spherical pyramid is a portion of the sphere bounded by a spherical polygon and the planes of the sides of the polygon. The centre of the sphere is the vertex of the pyramid, and the spherical polygon is its base. 696. Two spherical polygons are equal if they can be applied one to the other, so as to coincide. 697. Since the sides of a spherical polygon are arcs, they are usually expressed in degrees, minutes, and seconds. |