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PROP. XIV.

If a sphere be cut by a plane, the section will be a circle.

Because the radii of the sphere are all equal, each of them being equal to the radius of the describing semicircle, it is evident that if the section pass through the centre of the sphere, then the distance from the centre to every point in the periphery of that section will be equal to the radius of the sphere, and the section will therefore be a circle of the same radius as the sphere. But if the plane do not pass through the centre, draw a perpendicular to it from the centre, and draw any number of radii of the sphere to the intersection of its surface with the plane; then these radii are evidently the hypothenuses of a corresponding number of right-angled triangles, which have the perpendicular from the centre on the plane of the section, as a common side; consequently their other sides are all equal, and therefore the section of the sphere by the plane is a circle, whose centre is the point in which the perpendicular cuts the plane.

Cor. If two spheres intersect one another, the common section is a circle.

SCHOLIUM.

All the sections through the centre are equal to one another, and are greater than any other section which does not pass through the centre. Sections through the centre are called great circles, and the other sections small or less circles. Also, a straight line drawn through the centre of a circle of the sphere perpendicular to the plane of the circle is a diameter of the sphere, and the extremities of this diameter are called the poles of the circle. Hence it is evident that the arcs of great circles between the pole and circumference are equal, for the chords drawn in the sphere from either pole of a circle to the circumference are all equal.

PROP. XV.

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Every sphere is two-thirds of its circumscribing cylinder. Let ABCD be a cylinder circumscribing the sphere EFGH; then will the sphere EFGH be two-thirds of the cylinder ABCD. For let the plane AC be a section of the sphere and cylinder through the centre I, and join E AI, BI. Let FIH be parallel to AD or BC, and EIG and KL parallel to AB or DC, the base of the cylinder;' the latter line KL meeting BI in M, and the circular section of the sphere in N.

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Then, if the whole plane HFBC be conceived to revolve about the line HF as an axis, the square FG will describe a cylinder AG, and the quadrant IFG will describe a hemisphere EFG, and the triangle IFB will describe a cone IAB. Also, in the rotation, the three lines, or parts, KL, KN, KM, as radii, will describe corresponding circular sections of these solids, viz. KL a section of the cylinder, KN a section of the sphere, and KM a section of the

cone.

Now, FB being equal to FI, or IG, and KL parallel to FB, then by similar triangles IK=KM (Geom. Theor. 82), and IKN is a right-angled triangle; hence IN2 is equal to IK2+KN2 (Theor. 34). But KL is equal to the radius IG or IN, and KM=IK; therefore KL2 is equal to KM2+KN2, or the square of the longest radius of the said circular sections, is equal to the sum of the squares of the two others. Now circles are to each other as the squares of their diameters, or of their radii, therefore the circle described by KL is equal to both the circles described by KM and KN; or the section of the cylinder is equal to both the corresponding sections of the sphere and cone. And as this is always the case in every parallel position of KL, it follows that the cylinder EB, which is composed of all the former sections, is equal to the hemisphere EFG and cone IAB, which are composed of all the latter sections.

But the cone IAB is a third part of the cylinder EB (Prop. XIII. Cor. 2); consequently the hemisphere EFG is equal to the remaining two-thirds; or the whole sphere EFGH is equal to two-thirds of the whole cylinder ABCD.

Corol. I. A cone, hemisphere, and cylinder of the same base and altitude are to each other as the numbers 1, 2, 3.

Corol. 2. All spheres are to each other as the cubes of their diameters; all these being like parts of their circumscribing cylinders.

Corol. 3. From the foregoing demonstration it appears that the spherical zone or frustum EGNP is equal to the difference between the cylinder EGLO and the cone IMQ, all of the same common height IK. And that the spherical segment PFN is equal to the difference between the cylinder ABLO and the conic frustum AQMB, all of the same common altitude FK.

SCHOLIUM.

By the scholium to Prop. XIII. we have

cone AIB: cone QIM :: IF3: 1K3:: FH3:(FH—2FK)3 .. cone AIB: frustum ABMQ:: FH3: FH3—(FH—2FK)3 :: FH3:6FH2FK—12FH.FK2+8FK3;

but cone AIB = one-third of the cylinder ABGE; hence

cylinder AG: frustum ABMQ:: 3FH3:6FH2.FK-12FH.FK2+8FK3 Now cylinder AL: cylinder AG:: FK : FI

..

. cylinder AL: frustum ABMQ::6FH2: 6FH2-12FH.FK+8FK2

.. cylinder AL: segment

PFN::6FH2: 12FH.FK-8FK2, dividendo

::

FH2: FK(3FH-2FK).

But cylinder AL = circular base whose diameter is AB or FH multiplied by the height FK; hence cylinder AL= circle EFGH FK.

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SPHERICAL GEOMETRY.

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DEFINITIONS.

1. A SPHERE is a solid terminated by a curve surface, and is such that all the points of the surface are equally distant from an interior point, which is called the centre of the sphere.

We may conceive a sphere to be generated by the revolution of a semicircle APB about its diameter AB; for the surface described by the motion of the curve ABP will have all its points equally distant from the centre O

2. The radius of a sphere is a straight line drawn from the centre to any point on the surface.

The diameter or axis of a sphere is a straight line drawn through the centre, and terminated both ways by the surface.

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It appears from Def. 1, that all the radii of the same sphere are equal, and that all the diameters are equal, and each double of the radius.

3. It will be demonstrated, (Prop. 1.), that every section of a sphere, made by a plane, is a circle; this being assumed,

A great circle of a sphere is the section made by a plane passing through the centre of the sphere.

A small circle of a sphere is the section made by a plane which does not pass through the centre of the sphere.

4. The pole of a circle of a sphere is a point on the surface of the sphere equally distant from all the points in the circumference of that circle.

It will be seen, (Prop. 11.), that all circles, whether great or small, have two poles.

5. A spherical triangle is the portion of the surface of a sphere included by the arcs of three great circles.

6. These arcs are called the sides of the triangle, and each is supposed to be less than half of the circumference.

7. The angles of a spherical triangle are the angles contained between the planes in which the sides lie.

8. A plane is said to be a tangent to a sphere, when it contains only one point in common with the surface of the sphere.

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PROP. I.

Every section of a sphere made by a plane is a circle.
Let AZBX be a sphere whose centre is O.
Let XPZ be a section made by the plane XZ.
From O draw OC perpendicular to the plane XZ.
In XPZ take any points P1, P2, P3, . . . . .
Join CP1; CP2; CP3;
also, OP1; OP2 ;

OP 3;..

.....

Then, since OC is perpendicular to the plane XZ, it will be perpendicular to all straight lines passing through its foot in that plane. (Geometry of Planes.)

Hence, the angles OCP1, OCP2, OCP3, ·

C

B

3,....

are right angles,

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Hence, XPZ is a circle whose centre is C, and every other section of a sphere made by a plane may, in like manner, be proved to be a circle.

Cor. 1. If the plane pass through the centre of the sphere, then OC = o, and the radius of the circle will be equal to the radius of the sphere.

Cor. 2. Hence, all great circles are equal to one another, since the radius of each is equal to the radius of the sphere.

Cor. 3. Hence, also, two great circles always bisect each other, for their common intersection passing through the centre is a diameter.

Cor. 4. The centre of a small circle and that of the sphere, are in a straight line, which is perpendicular to the plane of the small circle.

Cor. 5. We can always draw one, and only one, great circle through any two points on the surface of a sphere, for the two given points and the centre of the sphere give three points, which determine the position of a plane.

If, however, the two given points are the extremities of a diameter, then these two points and the centre of the sphere are in the same straight line, and an infinite number of great circles may be drawn through the two points.

Distances on the surface of a sphere are measured by the arcs of great circles. The reason for this is, that the shortest line which can be drawn upon the surface of a sphere, between any two points, is the arc of a great circle joining them.

PROP. II.

If a diameter be drawn perpendicular to the plane of a great circle, the extremities of the diameter will be the poles of that circle, and of all the small circles whose planes are parallel to it.

Let APB be a great circle of the sphere whose centre is O.

Draw ZN a diameter perpendicular to the plane of circle APB.

Then, Z and N, the extremities of this diameter, are the poles of the great circle APB, and all the small circles, such as apb, whose planes are parallel to that of APB.

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PA

Pa

Through each of these points respectively, and the points Z and N, describe great circles, ZP,N, ZP,N.

Join OP1, OP 2,

Then, since ZO is perpendicular to the plane of APB, it is perpendicular to all the straight lines OP 1, OP 2,

drawn through its foot in that plane. ...... are right angles, and .. the

Hence, all the angles ZOP1, ZOP2, arcs ᏃᏢ), ZP 2,... 2,...... are quadrants. Thus, it appears that the points Z and N are equally distant from all the points in the circumference of APB, and are .. the poles of that great circle. Again, since ZO is perpendicular to the plane APB, it is also perpendicular to the plane apb, which is parallel to the former.

Hence, the oblique lines Zpı, Zp2,..... drawn to P1, P2, in the circumference of apb, will be equal to each other. (Geometry of Planes.) .. The chords Zp1, Zp2, ..... being equal, the arcs Zpı, Zp2, ·

which they subtend, will also be equal.

. The point Z is the pole of the circle apb; and, for the same reason, point N is also a pole.

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Cor. 1. Every arc P‚Z drawn from a point in the circumference of a great circle to its pole, is a quadrant, and this arc P、Z makes a right angle with the arc AP,B. For, the straight line ZO being perpendicular to the plane APB, `every plane which passes through this straight line will be perpendicular to the plane APB (Geometry of Planes); hence, the angle between these planes is a right angle, or, by (Def. 7), the angle of the arcs AP1 and ZP1 is a right angle.

Cor. 2. In order to find the pole of a given arc AP, of a great circle, take P1Z equal to a quadrant, and perpendicular to AP1, the point Z will be a pole of the arc AP1; or, from the points A and P, draw two arcs AZ and P、Z perpendicular to AP1, the point Z in which they meet is a pole of AP1.

Cor. 3. Reciprocally, if the distance of the point Z from each of the points A and P1 is equal to a quadrant, then the point Z is the pole of AP1, and each of the angles ZAP1, ZP,A, is a right angle.

For, let O be the centre of the sphere, draw the radii OA, OP1, OZ;

Then, since the angles AOZ, P、OZ, are right angles, the straight line OZ is perpendicular to the straight lines OA, OP1, and is . perpendicular to their

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