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PROPOSITION III. PROBLEM.

693. Given a material sphere to find its diameter.

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Let PBPC represent a material sphere.

It is required to find its diameter.

From any point P of the given surface, with any opening of the compasses, describe the circumference A B C on the surface.

Then the rectilinear line PB, being the opening of the compasses, is a known line.

Take any three points A, B and C in this circumference, and with the compasses measure the rectilinear distances AB, BC and CA.

Construct the AA'B'C', with its sides equal respectively to A B, BC and C A.

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Circumscribe a circle about the ▲ A'B'C'. The radius D'B' of this O is equal to the radius of O ABC. Construct the rt. Ab dp, having the hypotenuse b p =BP, and one side b d B' D'.

=

Draw bp to bp, and meeting p d produced in p'.

Then p p' is equal to the diameter of the given sphere.

For, if we bisect the sphere through P and B, and in the section draw the diameter PP' and chord BP', the ▲ bpp', when applied to ▲ BPP', will coincide with it.

Q. E. F.

PROPOSITION IV. THEOREM.

694. Through any four points not in the same plane, one spherical surface can be made to pass, and but one.

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Let A, B, C, D, be four points not in the same plane.

We are to prove that one, and only one, spherical surface can be made to pass through A, B, C, D.

Construct the tetrahedron A B CD, having for its vertices A, B, C, D.

Let E be the centre of the circle circumscribed about the face AB C.

Draw EM to this face.

Every point in EM is equally distant from the points A, B and C, § 450

(oblique lines drawn from a point to a plane at equal distances from the foot

of the are equal).

Also, let F be the centre of the circle circumscribed about

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... the plane passed through EH and FH is 1 to BC, § 449 (if a straight line be to two straight lines drawn through its foot in a plane, it is to the plane, and in this case the plane is to the line). to each of the faces A B C

Hence, this plane is also

and BCD,

§ 471

(if a straight line be to a plane, every plane passed through that line is to the plane).

.. the EM and FK lie in the plane EHF.

Hence they must meet unless they be parallel.

But if they were, the planes BCD and ABC would be one and the same plane, which is contrary to the hypothesis.

Now O, the point of intersection of the EM and FK, is equally distant from A, B and C; and also equally distant from B, C and D ;

.. it is equally distant from A, B, C and D.

Hence, a spherical surface, whose centre is 0, and radius O A, will pass through the four given points.

Only one spherical surface can be made to pass through the points A, B, C and D.

For the centre of such a spherical surface must lie in both the EM and FK.

And, since is the only point common to theses, O is the centre of the only spherical surface passing through A, B, C and D.

Q. E. D.

695. COROLLARY 1. The four perpendiculars erected at the centres of the faces of a tetrahedron meet at the same point.

696. COR. 2. The six planes perpendicular to the six edges of a tetrahedron at their middle point will intersect at the same

PROPOSITION V. THEOREM.

697. A sphere may be inscribed in any given tetrahedron.

D

A

B

Let ABCD be the given tetrahedron.

We are to prove that a sphere may be inscribed in ABC D. Bisect the dihedral at the edges A B, BC and AC by the planes OA B, OBC and OA C respectively.

Every point in the plane OA B is equally distant from the faces ABC and A B D,

§ 477 For a like reason, every point in the plane OBC is equally distant from the faces A B C and D BC;

and every point in the plane OA C is equally distant from the faces A B C and A DC.

.. O, the common intersection of these three planes, is equally distant from the four faces of the tetrahedron.

.. a sphere described with O as a centre, and with the radius equal to the distance of O to any face, will be tangent to each face, and will be inscribed in the tetrahedron.

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Q. E. D.

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

ON DISTANCES MEASURED ON THE SURFACE OF THE SPHERE.

699. DEF. The distance between two points on the surface of a sphere is understood to be the arc of a great circle joining the points, unless otherwise stated.

700. DEF. The distance from the pole of a circle to any point in the circumference of the circle is the Polar distance of the circle.

PROPOSITION VI. THEOREM.

701. The distances measured on the surface of a sphere from all points in the circumference of a circle to its pole are

equal.

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Let P, P' be the poles of the circle A BC.

We are to prove arcs P A, PB, PC equal.

§ 450

The rectilinear lines PA, P B and PC are equal, (oblique lines drawn from a point to a plane at equal distances from the foot

of the

are equal);

.. the arcs P A, P B and P C.are equal,

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(in equal

equal chords subtend equal arcs).

In like manner arcs P'A, P' B and P' C are equal.

Q. E. D.

702. COROLLARY 1. The polar distance of a great circle is a quadrant. Thus, arcs PA', PB', P' A', P' B', polar distances of the great circle A' B' C' D', are quadrants; for they are the measures of the right angles A' O P, B'OP, A'O P', B' O P', whose vertices are at the centres of the great circles PA' P'C', PB' P' D'.

703. COR. 2. If a point P on the surface of the sphere be at the distance of a quadrant from the two points A' and B' of an arc of a great circle, it is the pole of that arc. For, since the arcs PA' and P B' are quadrants, the angles PO A' and PO B' are right angles. And hence, the radius O P is perpendicular to each of the lines OA' and O B', and therefore perpendicular to the plane of the arc A'B' (§ 449). Therefore P is the pole of arc A'B'.

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