SOLUTION.-Let A and B be two points on the surface of a sphere, through which it is proposed to pass a circumference of a great circle. From B as a pole, with an arc equal to a quadrant, strike an arc on, as nearly where the pole of the circle passing through A and B lies, as may be determined by inspection. Then, from A, with the same arc, strike an arc st intersecting on at P. Now, P is the pole of the great circle passing through A and B. Hence, from P as a pole, with a quadrant arc draw a circle; it will pass through A and B, and will be a great circle, since its pole is a quadrant's distance from its circumference. [The student should make the construction on the spherical blackboard.]

FIG. 316.

549. COR. 1.-Through any two points on the surface of a sphere, one great circle* can always be made to pass, and only one, except when the two points are at the extremities of the same diameter, in which case an infinite number of great circles can be passed through the two points.

Since the arcs on and st are arcs of great circles, the circumferences of which they form parts will intersect also on the opposite side of the sphere, at a distance of a semicircumference from P. But these two points are poles of the same great circle. Now, as the two great circles can intersect at no other points, there can be only one great circle passed through A and B. But if the two given points were at the extremities of the same diameter, as at D and C, the arcs st and on would coincide, and any point in this circumference being taken as a pole, great circles can be drawn through D and C. [The student should trace the work on the spherical blackboard.]

550, SCH.-The truth of the corollary is also evident from the fact that three points not in the same straight line determine the position of a plane. Thus A, B, and the centre of the sphere, fix the position of one, and only one, great circle passing through A and B. Moreover, if the two given points are at the extremities of the same diameter, they are in the same straight line with the centre of the sphere, whence an infinite number of planes can be passed through them and the centre. The meridians on the earth's surface afford an example, the poles (of the equator) being the given points.

551. Cor. 2.—If two points in the circumference of a great circle of a sphere, not at the extremities of the same diameter, are at a quadrant's distance from a point on the surface, that point is the pole of the circle.

* The word circle may be understood to refer either to the circle proper, or to its circumference. The word is in constant use in the higher mathematics, in the latter sense.

PROPOSITION VI.

552. Theorem.-The shortest distance on the surface of a sphere, between any two points in that surface, is measured on the arc less than a semicircumference of the great circle which joins them.

m

FIG. 317.

DEM.-Let A and B be any two points in the surface of a sphere, AB the arc of a great circle joining them, and AmCnB any other path in the surface between A and B; then is arc AB less than AmCnB.

Let C be any point in AmCnB, and pass the arcs of great circles through A and C, and B and C. Join A, B, and C with the centre of the sphere. The angles AOB, AOC, and COB form the facial angles of a triedral, of which angles the arcs AB, AC, and CB are the measures. Now, angle AOB < AOC + COB (434); whence arc AB <arc AC + arc CB, and the path from A to B is less on arc AB than on arcs AC, CB. In like manner, joining any point in AmC with A and C by arcs of great circles, their sum would be greater than AC. So, also, joining any point in CB with C and B, the sum of the arcs would be greater than CB. As this process is indefinitely repeated, the path from A to B on the arcs of the great circles will continually increase, and also continually approximate the path AmCnB. Hence, arc AB is less than the

path AmCnB. Q. E. D.

553, COR.-The least arc of a circle of a sphere joining any two points in the surface, is the arc less than a semicircumference of the great circle passing through the points; and the greatest arc is the circumference minus this least arc.

A

m

DEM.-Let AmBn be any small circle passing through A and B, and ABDoC the great circle. As shown above, ApB < AmB. Now, circumference ABDOC > circumference AmBn (539). Subtracting the former inequality from the latter, we have BDOCA BRA. Q. E. D.

PROPOSITION VII.

554. Theorem.-The shortest path on the surface of a hemisphere, from any point therein to the circumference of the great circle forming its base, is the arc less than a quadrant of a great circle perpendicular to the base, and the longest path, on any arc of a great circle, is the supplement of this shortest path.

DEM.-Let P be a point in the surface of the hemisphere whose base is ACBC', and DPmD' an arc of a great circle passing through P and perpendicular to ADCBC'; then is PD the shortest path on the surface from P to circumference ADBC', and PmD' is the longest path from P to the circumference, measured on the arc of a great circle.

n

m

FIG. 319.

B

For, the shortest path from P to any point in circumference ADBC' is measured on the arc of a great circle (552). Now, let PC be any oblique arc of a great circle. We will show that are PD < arc PC. Produce PD until DP′ = PD; and pass a great circle through P' and C. Draw the radii OP, OD, OC, and OP'. The triedrals O-PDC and O-P'DC have the facial angle POD= P'OD, they being measured by equal arcs, and the facial angle DOC common. Hence, as the included diedrals are equal, both being right, the triedrals are equal or symmetrical (446). In this case they are symmetrical, and the facial angle POC = P'OC; whence the arc PC arc P'C. Finally, since PC + P'C > PP', PC, the half of PC + P'C, is greater than PD, the half of PP'.

=

Secondly, PmD' is the supplement of PD, and we are to show that it is greater than any other arc of a great circle from P to the circumference ADBC'. Let PnC' be any arc of a great circle oblique to ADCBC'. Produce C'nP to C. Now CPNC' is a semicircumference and consequently equal to DPmD'. But we have before shown that PD < PC, and subtracting these from the equals CPC' and DPmD', we have PmD'> PnC'.

555. COR.-From any point in the surface of a hemisphere there are two perpendiculars to the circumference of the great circle which forms the base of the hemisphere; one of which perpendiculars measures the least distance to that circumference, and the other the greatest, on the arc of any great circle of the sphere.

Thus PD and PmD' are two perpendiculars from P upon the circumference ADBC'.

FIG. 321.

B

B'

557. A Spherical Angle is the angle included by two arcs of great circles.

ILL.-BAC, Fig. 321, is a spherical angle, and is conceived as the same as the angle B'AC', B'A and C'A being tangents to the great circles BADF and CAEF. [The student should not confound such an angle as BAC, Fig. 320, with a spherical angle.]

PROPOSITION VIII.

558. Theorem.-A spherical angle is equal to the measure of the diedral included by the great circles whose arcs form the sides of the angle.

DEM.-Let BAC be any spherical angle, and BADF and CAEF the great circles whose arcs BA and CA include the angle; then is BAC equal to the measure of the diedral C-AF-B. For, since two great circles intersect in a diameter (542), AF is a diameter. Now B'A is a tangent to the circle BADF, that is, it lies in the same plane and is perpendicular to AO at A. In like manner C'A lies in the plane CAEF and is perpendicular to AO. Hence B'AC' is the measure of the diedral C-AF-B (425). Therefore the spherical angle BAC, which is the same as the plane angle B'AC', is equal to the measure of the diedral C-AF-B. Q. E. D.

FIG. 322.

559. COR. 1.-If one of two great circles passes through the pole of the other, their circumferences intersect at right angles.

A

m

FIG. 323.

B

DEM. Thus, P being the pole of the great circle CABM, PO is its axis, and any plane passing through PO is perpendicular to the plane CABm (429).~ Hence, the diedral B-AO-P is right, and the spherical angle PAB, which is equal to the measure of the diedral, is also right.

560. Cor. 2.-A spherical angle is measured by the arc of a great circle intercepted between its sides, and at a quadrant's distance from its vertex.

Thus, the spherical angle CPA is measured by CA, PC and PA being quadrants. For, since PC is a quadrant, CO is perpendicular to PO, the edge of the diedral C-PO-A, and for a like reason AO is perpendicular to PO. Hence, COA is the measure of the diedral, and consequently CA, its measure, is the measure of the spherical angle CPA.

561. COR. 3.-The angle included by two arcs of small circles is the same as the angle included by two arcs of great circles passing through the vertex and having the same tangents.

Thus BAC B"AC". For the angle BAC is, by definition, the same as B'AC', B'A and C'A being tangents to BA and CA. Now, passing planes through C'A, B'A, and the centre of the sphere, we have the arcs B'A, C"A, and B'A, C'A tangents to them. Hence, B"AC" is the same as B'AC', and consequently the same as BAC.

FIG. 324.

B"

562. SCH.-To draw an arc of a great circle which shall be perpendicular to another; or, what is the same thing, to construct a right spherical angle. Let it be required to erect an arc of a great circle perpendicular to CAB at A, Fig. 323. Lay off from A, on the arc CAB, a quadrant's distance, as AP', and from P' as a pole, with a quadrant describe an arc passing through A. This will be the perpendicular required. In a similar manner we may let fall a perpendicular from any point in the surface, upon any arc of a great circle. To let fall a perpendicular from P upon the arc CAB, from P as a pole, with a quadrant describe an arc cutting CAB, as at P'. Then from P' as a pole, with a quadrant describe an arc passing through P and cutting CAB, and it will be perpendicular to CAB. [The student should have practice in making these constructions on the sphere.]

PROPOSITION IX.

563. Problem.-To pass the circumference of a small circle through any three points on the surface of a sphere.

a.

SOLUTION.-Let A, B, and C be the three points in the surface of the sphere through which we propose to pass the circumference of a circle. Pass arcs of great circles through the points, forming the spherical triangle ABC. Thus, to pass an arc of a great circle through B and C, from B as a pole, with a quadrant strike an arc as near as may be to the pole of the required circle; and from C as a pole, with the quadrant strike an arc intersecting the former, as at P; then is P the pole of a great circle passing through B and C (?). Hence, from P as a pole, with a quadrant pass an arc through B and C, and it will be the arc required (551). In like manner pass arcs through A and C, A and B. Now, bisect two of these arcs, as BC and AC, by arcs of great

FIG. 925.