ON THE QUADRATURE OF THE CIRCLE, AND THE RECTIFICATION OF ITS CIRCUMFERENCE.

DEFINITIONS.

1. The determination of a square equal to a given surface, is called its quadrature.

2. The determination of a straight line equal to a curve line, is called its rectification.

3. A mixed line is composed of straight and curve lines. 4. A mixtilineal space is a space contained by a mixed line.

5. The inclination of a straight line and a curve is the angle contained by the former, and a tangent to the latter at the point of intersection.

6. The supplemental chord of an arc is the chord of its defect from a semicircumference.

AXIOMS.

1. Of all lines that can join two points, there must be at least one such that no other is less than it; and if there be only one such, it is the least.

2. If any number of lines fulfil certain conditions, and if a line fulfilling the same conditions can always be found less than any of them, except one, this one must be the least.

If two rectilineal figures be on the same side of the same base, and if one of them be wholly encompassed by the other, and be also concave internally, the sum of its sides is less than that of the other.

Let the figures ACDB, AEFB, be upon the same base AB, then AE + EF + FB AC + CD + DB.

G

F

G

K

For, produce AC to G; join C, F, and produce CD to K; then (I. 20) AE + EG — AG, CG+GFCF, CF+FK7CK, E and DK+KBDB. Therefore AE+ EF +FB AG + GF + FB, which is AC+ CF+FB, which is AC + CK +KB, which is AC+ CD + DB, or AE+EF+FB AC + CD + DB.

A

B

And, in a similar manner, the proposition is proved for

polygons.

Of all the lines, straight and curve, that can join two points, the straight line is the least.

Let A, B, be any two points, the straight line AB is the shortest line that can join them.

1. Let the curve line ADB, concave towards AB, join them. Then take any point C in AB, and from the centres A and B, with the radii AC, CB, respectively, describe the arcs CD, CE, passing through C, and cutting ADB in D and E. Then, since the A

D E

C

arcs touch at C (III. 12), and cannot touch at any other point (III. 13), the point E must be without the arc CD, and D without the arc CE; and the line DE between them must be of some length. Now, if the points D and E be made to coincide with C, A and B remaining fixed, the curve lines AD, EB, would connect A and B; and hence a line shorter than ADEB can connect these points. The same can be proved of any other concave line, therefore the straight line AB is shorter than any of them (Ax. 1).

C

2. Let a line ACDB, partly concave and partly convex towards AB, connect the points. Let AC, DB, be concave, and CD the convex portion. Draw the straight lines AC, CD, DB. Then (by 1st case) the straight lines AC, CD, DB, are less than the curve lines A

AC, CD, DB, respectively; and the sum of the former therefore less than the whole curve line; hence a line shorter than it has been found.

Then, if AD, and

3. Let the crooked line ACDB join A and B. AD be joined by a straight line, AC + CD AD+DBAB; therefore the sum of the straight lines AC, CD, and DB, are greater than AB.

Since, therefore, a line shorter than any of the curve or crooked lines that join A and B can be found, but none shorter than the straight line can be found, therefore it is the shortest (Ax. 1).

D

COR. 1.-If two points A, B, be joined by a straight line AB, and a curve line ACB, or a mixed line, which is concave towards AB; ACB is the least of all the lines that lie on the same A

C

B

side of AB, that join A and B, and that do not lie between ACB and AB.

For if ADB be any other line, then by joining two points in it by a straight line, a shorter line will be found than it, joining A and B, and fulfilling the other conditions; and this will be the case whether ADB be composed of straight lines, or be a mixed line, or a curve line. And as this can be proved of every line fulfilling the required conditions, except of ACB; therefore (Ax. 2) ACB is the least.

COR. 2.-If two intersecting curves, whose curvatures lie in one and the same direction, be cut by a straight line, inclined to the interior at an angle not greater than a right angle, the arc which it intercepts on this line from the point of intersection, will be less than the corresponding arc of the other line.

Let AB, AC, intersect in A, and be cut by CD, so that the angle at B is less than a right angle, then AC AB.

For, if CE, BE, be curves equal to AC, AB, in every respect, on the other side of CD, but in a reverse position, and having respectively A

B

D

E

the same inclinations to CD; then, since the angles at B do not exceed two right angles, ABE is concave towards D, and hence (last Cor.) ACE ABE, and therefore AC AB.

COR. 3.-Hence the circumference of a circle is greater

than the perimeter of any inscribed polygon, and less than that of any one circumscribing it.

PROPOSITION III. THEOREM.

If from the greater of two unequal magnitudes there be taken away its half, and from the remainder its half, and so on, there shall at length remain a magnitude less than the least of the proposed magnitudes.

Let AB and C be two unequal magnitudes, of which AB is the greater. If from AB there be taken away its half, and from the remainder its half, and so on, there shall at length remain a magnitude less than C.

A

For C may be multiplied so as at length to become greater than AB. Let it be so multiplied, and let DE its multiple be greater than AB, and let DE be divided into DF, FG, GE, each equal to C. From AB take BH equal to its half, and from the remainder AH take HK equal to its half, and so on, until there be as many divisions in AB as there are in DE; and let the divisions in AB be AK, KH, HB; and the divisions in ED be DF, FG, GE; and because DE is greater than AB, and that EG taken from DE is not greater than its half, but BH taken from AB is equal to its half; therefore the remainder GD is greater than the remainder HA. Again, because GD is greater B E than HA, and that GF is not greater than the half of GD, but HK is equal to the half of HA; therefore the remainder FD is greater than the remainder AK; and FD is equal to C; therefore C is greater than AK; that is, AK is less than C.

PROPOSITION IV. THEOREM.

Equilateral polygons of the same number of sides inscribed in circles, are similar, and are to one another as the squares of the diameters of the circles.

Let ABCDEF and GHIKLM be two equilateral polygons of the same number of sides inscribed in the circles ABD and GHK; ABCDEF and GHIKLM are similar, and are to one another as the squares of the diameters of the circles ABD, GHK.

Find N and O the centres of the circles; join AN and BN, as also GO and HO, and produce AN and GO till they meet the circumferences in D and K.

Because the chords AB,

BC, CD, DE, EF, FA, are all equal, the arcs AB, BC, CD, DE, EF, FA, are also equal (III. 28); for the same reason, the arcs GH,

HI, IK, KL, LM, MG, are all equal, and they are equal in number to the others; therefore, whatever part the arc AB is of the whole circumference ABD, the same is the arc GH of the circumference GHK ; but the angle ANB is the same part of four right angles that the arc AB is of the circumference ABD (VI. 33, Cor. 2); and the angle GOH is the same part of four right angles that the arc GH is of the circumference GHK; therefore the angles ANB, GOH, are each of them the same part of four right angles, and therefore they are equal to one another. The isosceles triangles ANB, GOH, are therefore equiangular (VI. 6), and the angle ABN equal to the angle GHO; in the same manner, by joining NC, OI, it may be proved that the angles NBC, OHI, are equal to one another, and to the angle ABN; therefore the whole angle ABC is equal to the whole GHK; and the same may be proved of the angles BCD, HIK, and of the rest; therefore the polygons ABCDEF and GHIKLM are equiangular to one another; and since they are equilateral, the sides about the equal' angles are proportionals; the polygon ABCDEF is therefore similar to the polygon GHIKLM; and because similar polygons are as the squares of their homologous sides (VI. 20), the polygon ABCDEF is to the polygon GHIKLM as the square of AB to the square of GH; but because the triangles ANB, GOH, are equiangular, the square of AB is to the square of GH, as the square of AN to the square of GO (VI. 4, and 22, Cor.), or as four times square of AN to four times the square of GO, that is, as the square of AD to the square of GK; therefore, also, the polygon ABCDEF is to the polygon GHIKLM as the square of AD to the square of GK; and they have also been shown to be similar.

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