Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

3 P. The side of any equilateral polygon inscribed in a circle being given, to find the side of a polygon of the same number of sides described about the circle.

Let ABCDEF be an equilateral polygon inscribed in the circle ABD; it is required to find the side of an equilateral polygon of as many sides described about the circle.

H Find G the centre of the circle; join GA, GB; bisect the arc AB in H; and through H draw LK touching it the circle, and meeting GA, GB produced in K and L; KL is the side of the polygon required.

Produce GF to Ñ, so that GN may be equal to GL; join KN, and from Ġ draw GM at right angles to KN, join also HG.

Because the arc AB is bisected in H, the angles AGH, BGH are equal (p. 27 of b. 3); and because KL touches the circle in H, the angles GHL, GHK are right angles (p. 18 of b. 3); therefore, the triangles HGK, HGL, have two angles in the one equal to two angles in the other, and the side GH is common to both; therefore they are equal (p. 26 of b. 1), and GL is equal to GK.

Again, in the triangles KGL, KGN, because GN is equal to GL, and GK common, and also the angles KGL, KGN equal ; therefore the bases KL and KN are equal (p. 4 of b. 1). But because the triangle KGN is isosceles, the angles GKN and GNK are equal; and the angles GMK, GMN were made right angles; wherefore the triangles GMK, GMN have two angles in the one equal to two in the other, and the side GM is common to both; therefore they are equal (p. 26 of b. 1), and KN is bisected in M. But KN is equal to KL, therefore their halves KM, KH are also equal. Wherefore, in the triangles GKH, GKM, the two sides GK and KH are equal to the two GK and KM, each to each; the angles GKH, GKM are also equal; therefore GM is equal to GH (p. 4 of b. 1). Wherefore the point M is in the circumference of the circle ; and because KMG is a right angle, KM touches the circle. Therefore KL is the side of an equilateral polygon described about the circle, of as many sides as the inscribed polygon ABCDEF.

Cor. 1. Because of the equal straight lines GL, GK, GN, if a circle be described from the centre G, through the points L, K, N, the polygon will be inscribed in that circle; and be similar to the polygon ABCDEF.

Cor. 2. The sides of the inscribed and described polygons, have to each other the ratio of the perpendiculars let fall from G upon AB and LK. Therefore because magnitudes have the ratio of their equimultiples (p. 15 of b. 5), the perimeters which are equimultiples of the sides of the polygons, are to each other as the perpendicular from the centre upon a side of the inscribed polygon is to the radius of the circle.

B

KAM

B

מ

4 Th. A circle being given, two similar polygons may be found, the one inscribed in the circle, the other described about it, whose difference shall be less than any given space. Let ABC be the given circle, and

LF the square D any given space; a polygon may be inscribed in the circle ABC, and a similar one described HVA about it, whose difference shall be less than the square of D.

In the circle ABC, apply the straight line AE, equal to D, and let AB be a quadrant." From AB take its half, from the residue its half, and so on, until the arc AF, be less than the arc AE (p. 1 of this). Find the centre G, draw the diameter AC, as also the straight lines AF, FG: bisecting the arc AF in K, join KG; and draw HL touching the circle in K, and meeting GA, GF produced in H and L: join CF.

Because the isosceles triangles HGL, AGF have the common angle AGF, they are equiangular (p: 6 of b. 6); therefore the angles, GHK, GAF are equal; but the angles GKH, CFA are also equal as right angles; therefore the triangles HGK, ACF, are likewise equiangular (p. 32 of b. 1).

And because the arc AF was found by taking from the arc AB its half, and from the residue its half, and so on; Af will be found a certain number of times exactly in AB, and therefore also in the whole circumference ABC. AF is therefore the side of an equilateral polygon inscribed in the circle ABC. Wherefore also, HL is the side of an equilateral polygon of the same number of sides described about the circle ABC (p. 3 of this).

Let the polygon described be called M, and the polygon inscribed be called Ñ; and because they are similar, they are as the squares of the homologous sides HL and AF, (p. 20 of b. 6), that is (because the triangles HLG, AFG, are similar), as the square of HG to the square of AĞ, or its equal GK. But the triangles HGK, ACF were shown to be similar, and therefore the square of AC is to the square of CF, as the polygon M to the polygon N; and, by conversion, the square of AC is to its excess above the square of CF, viz. the square of AF (p. 47 of b. 1), as the polygon M is to its excess above the polygon N. But the square of AC, which is about the circle ABC, is greater than the regular octagon which is about the same circle; because the square envelopes the octagon (Lemma); and for the same reason, the polygon of eight sides is greater than one of sixteen, and so on: therefore the square of AC is greater than any polygon described about the circle by the continual bisection of the arc ABC: it is therefore greater than the polygon M.

Now, it has been demonstrated, that the square of AC is to the square of AF, as the polygon M to the difference of the polygons ; therefore, since the square of AC is greater than M, the square of AF is greater than the difference of the polygons, (p. 14 of 6. 5). The difference of the polygons is therefore less than the square of ÁF; but

AF is less than D; therefore, the difference of the polygons is less than the square of D; that is, than the given space.

Cor. 1. Because the polygons M and N differ from each other more than either of them differs from the circle, the difference between each of them and the circle is less than the space given, the square of D. Therefore, however small a given space may be, a polygon may be inscribed in the circle, and another scribed about it, each of which shall differ from the circle by a space less than the given space.

Cor 2. The space B, which is greater than any polygon that can be inscribed in the circle A, and less than any polygon that can be described about it, is equal to the circle A. "If not equal, let B exceed the circle A, by the space C. Then, because the polygons de scribed about the circle A are all greater than D, by hypothesis; and because B is greater than A by the space C, therefore no polygon can be described about the circle A, that does not exceed it by a space greater than C; which is absurd. In the same manner, if B be less than A by the space C, it is shown that no polygon can be inscribed in the circle A, that is not less than A, by a space greater than C, which is also sbsurd. Therefore A and B are not unequal; that is, they are equal to one another,

[ocr errors]

M

5 Th. The area of any circle is equal to the rectangle contained by the radius and a straight line equal to half the circumference.

Let ABC be a circle, of which the centre is D, and the diameter AC; if in AC produced, there be taken AH

Napo equal to half the circumference, the area of the circle is equal to the rect- EA

KHL angle contained by DA and AH.

Let AB be the side of any equilateral polygon inscribed in the circle ABC; bisect the arc AB in G, and through G draw EF touching the circle, and meeting DA produced in E, and DB produced in F, EF will be the side of an equilateral polygon described about the circle ABC (p. 3 of this). In AC produced take AK equal to half the perimeter of the polygon whose side is AB; and AL equal to half the perimeter of the polygon whose side is EF. Then AK will be less, and AL greater than the straight line AH. (Lemma of this)Now, because in the triangle EDF, DG is drawn perpendicular to the base, the triangle EDF is equal to the rectangle of DG and the half of EF, (p. 41 of b. 1), and as the same is true of all the triangles having their vertices in D, which make up the polygon described about the circle ; therefore the whole polygon is equal to the rectangle contained by DG or DA and AL (p. 1 of b. 2). But AL is greater than AH; therefore the rectangle DAXAL is greater than the rectangle DAXAH; therefore the area of the circle ABC is less than that of any polygon deseribed about it.

Again, the triangle ADB is equal to the rectangle contained by DM,

the perpendicular, and half the base AB; and it is therefore less than the rectangle contained by DG or DA and the half of AB. And as the same is true of all the other triangles having their vertices in D, which make up the inscribed polygon; therefore the whole of the inscribed polygon is less than the rectangle contained by DA and AK-half the perimeter of the polygon. Now the rectangle DA, AK is less than the rectangle DA, AH, and still less is the polygon whose side is AB. The rectangle DA, AH is therefore greater than any polygon inscribed in the circle ABC: and the same rectangle has been proved to be less than any polygon described

about the circle; therefore the rectangle of DA, the radius, and AH, half the circumference, is equal to the area of the circle ABC. (Cor. 2 p. 4 of this.)

Cor. 1. Because DA: AH :: DA2 : DAXAH (p. 1 of b. 6); and that DAXAH= area of the circle of which DA is the radius : therefore, as the radius of any circle is to the semi-circumference; or as the diameter to the circumference, so is the square of the radius to the area of the circle.

Cor. 2. Hence a polygon may be described about a circle, the perimeter of which shall exceed the circumference of the circle by a line that is less than any given line. Let No be the given line. Take in NO the part NP less than its half, and less than AĎ; and let a polygon be described about the circle ABC, so that its excess above ABC may be less than the square of NP (cor. 1, p. 4 of this). Let the side of this polygon be EF. And since, as above, the circle is equal to the rectangle DA, AH, and the polygon to the rectangle DA, AL, the excess of the polygon above the circle is equal to the rectangle DA, HL; therefore the rectangle DA, HL is less than the square of NP; and therefore, since DA is greater than NP, HL is less than NP, and twice HL less than twice NP, wherefore twice HL is still less than NO. But HL is the difference between half the perimeter of the polygon whose side is EF, and the semi-circumference of the circle; therefore twice HL is the difference between the perimeter of the polygon and the circumference of the circle (p. 5 of b. 5); which is therefore less than NO.

Cor. 3. Hence also a polygon may be inscribed in a circle, such that the excess of the circumference above the perimeter of the polygon, may be less than any given line. This may be proved like the preceding.

6 Th. The areas of circles are in the duplicate ratio ; or as the squares of their diameters.

Let ABD and GHL be two circles, of which the diameters are AD and GL; the

R circıe ABD is to the circle GHL as the square of AD to the

Y square of GL.

Let ABCDEF and GHKLMN be two equilateral polygons

H Н

[ocr errors]

M

of the same number of sides inscribed in the circles ABD, GHL; and let Q be such a space, that the square of AD is to the square of GL as the circle ABD is to the space Q.

Because the polygons ABCDEF and GHKLMN are equilateral and of the same number of sides they are similar (p. 2 of this), and their areas are as the squares of the diameters of the circles in which they are inscribed. Therefore ADP : GL? :: polygon ABCDEF :: polygon GHKLMN ;-but AD2 : GL? : : circle ABD : Q. Now the circle ABD is greater than the polygon ABCDEF; therefore Q is greater than the polygon GHKLMN (p. 14 of b. 5); that is, Q is greater than any polygon inscribed in the circle GHL.

к

In the same manner it is demonstrated, that Q is less than any polygon described about

R the circle GHL ; wherefore the space Q is equal to the circle GHL (cor. 2, p. 4 of this).

P Now, by the hypothesis, the circle ABD is to the space Q as the square of AD to the square of GL; therefore the circle ABD is to the circle GHL, as the square of AD is to the

M

square of GL.

Cor. 1. Hence the circumferences of circles are to each other as their diameters.

Let the straight line X be equal to half the circumference of the circle ABD, and the straight line Y be equal to half the circumference of the circle GHL: and because the rectangle AOXX and GPXY are equal to the circles ABD, GHL (p. 5 of this); therefore AOXX : GPXY :: AD? : GL’, and as A02 ; GP?—and alternately, AOXX : A0?. : : GPXY : GP2. Whence, because rectangles that have equal altitudes are as their bases (p. 1 of b. 6), X is to AO as Y is to GP, and alternately X:Y::A0: GP. Wherefore, taking the doubles of each, the circumference ABD is to the circumference ĞHL, as the diameter AD to the diameter GL.

Cor. 2. The circle that is upon the side of a triangle opposite the right angle, is equal to the two circles described upon the sides containing the righi angle.

For the circle described upon SR is to the circle described upon RT as the square of SR to the square of RT; and the circle described upon TS is to the circle described upon RT, as the square of ST to the square of RT.

Wherefore, the circles described upon SR and on ST are to the circle described on RT, as the squares of SR and of ST to the square of RT (p. 24 of b. 5). But the squares of RS and ST are equal to the square of RT (p. 47 of b. 1); therefore the circles described on RS and ST are equal to the circle described on RT.

« ΠροηγούμενηΣυνέχεια »