Let MN be equal to the given difference of the adjacent sides of the required rectangle; and let the given area be that of the square described on AB.

=

Upon MN as a diameter describe a circle. At M draw the tangent MP AB, and from P, draw the secant PQR through the centre of the circle; then, PR and PQ are the base and

=

altitude of the required rectangle. For, by (III. 59), PR × PQ = AB', and the difference of PR and PQ is QR = MN.

PM2

PROPOSITION XVII.-PROBLEM.

37. To find two straight lines in the ratio of the areas of two given polygons.

C

D

B

Let squares be found equal in area to the given polygons, respectively (30). Upon the sides of the right angle ACB, take CA and CB equal to the sides of these squares, join AB and let fall CD perpendicular to AB. Then, by (III. 46), we have ᎪᎠ : ᎠᏴ == CA: CB2; therefore, AD, DB, are in the ratio of

the areas of the given polygons.

PROPOSITION XVIII.-PROBLEM.

38. To find a square which shall be to a given square in the ratio of

ference in F; join FC, FE; lay off FH = AB, and through H draw HG parallel to EC; then, FG is the side of the required square. For, by (III. 15), we have

But FH = AB, therefore the square constructed upon FG is to the square upon AB in the ratio M: N.

PROPOSITION XIX.-PROBLEM.

39. To construct a polygon similar to a given polygon and whose area shall be in a given ratio to that of the given polygon.

Let P be the given polygon, and let a be one of its sides; let M: N be the given ratio.

Find, by the preceding problem, the side a' of a square which shall be to a2 in the ratio M: N; upon a', as a homologous side to a, construct the polygon P' similar to P (III. 80); this will be the polygon required.

For, the polygons being similar, their areas are in the ratio a'2: a2, or M: N, as required.

Mr

N

P

a

PROPOSITION XX.-PROBLEM.

40. To construct a polygon similar to a given polygon P and equivalent to a given polygon Q.

Find M and N, the sides of squares respectively equal in area to P and Q, (30).

Let a be any side of P, and find a fourth proportional a' to M, N and a: upon a', as a homologous side to a, construct the polygon P' similar to P; this will be the required polygon. For, by construction,

P

Q

a

M

N

therefore, taking the letters P, Q and P', to denote the areas of the

but, the polygons P and P' being similar, we have, by (23),

and comparing these equations, we have P' = Q.

Therefore, the polygon P' is similar to the polygon P and equivalent to the polygon Q, as required.

REGULAR POLYGONS.

BOOK V.

MEASUREMENT OF THE CIRCLE.

MAXIMA AND MINIMA OF PLANE FIGURES.

REGULAR POLYGONS.

1. DEFINITION. A regular polygon is a polygon which is at once equilateral and equiangular.

The equilateral triangle and the square are simple examples of regular polygons. The following theorem establishes the possibility of regular polygons of any number of sides.

PROPOSITION I.-THEOREM.

G

2. If the circumference of a circle be divided into any number of equal parts, the chords joining the successive points of division form a regular polygon inscribed in the circle; and the tangents drawn at the points of division form a regular polygon circumscribed about the circle. Let the circumference be divided into the equal arcs AB, BC, CD, etc.; then, 1st, drawing the chords AB, BC, CD, etc., ABCD, etc., is a regular inscribed polygon. For, its sides are equal, being chords of equal arcs; and its angles are equal, being inscribed in equal segments.

2d. Drawing tangents at A, B, C, etc., the polygon GHK, etc., is a regular circumscribed

H

C

K

polygon. For, in the triangles AGB, BHC, CKD, etc., we have AB BC CD, etc., and the angles GAB, GBA, HBC, HCB,

=

etc., are equal, since each is formed by a tangent and chord and is measured by half of one of the equal parts of the circumference

(II. 62); therefore, these triangles are all isosceles and equal to each other. Hence, we have the angles G

GB BH = HC = CK, etc., from equals, it follows that GH = HK, etc.

=

H = K, etc., and AG=

which, by the addition of

3. Corollary I. Hence, if an inscribed polygon is given, a circumscribed polygon of the same number of sides can be formed by drawing tangents at the vertices of the given polygon. And if a circumscribed polygon is given, an inscribed polygon of the same number of sides can be formed by joining the points at which the sides of the given polygon touch the circle.

It is often preferable, however, to obtain the circumscribed polygon from the inscribed, and reciprocally, by the following methods:

B'

F C'

B

G

D

1st. Let ABCD.... be a given inscribed polygon. Bisect the arcs AB, BC, CD, etc., in the points E, F, G, etc., and draw tangents, A'B', B'C', C'D', etc., at these points; then, since the arcs EF, FG, etc., are equal, the polygon A'B'C'D'.... is, by the preceding propoition, a regular circumscribed polygon of the same number of sides as ABCD.... Since the radius OE is perpendicular to

A'

A

AB (II. 16) as well as to A'B', the sides A'B', AB, are parallel; and, for the same reason, all the sides of A'B'C'D' .... are parallel to the sides of ABCD.... respectively. Moreover, the radii OA, OB, OC, etc., when produced, pass through the vertices A,B,C', for since B'E : B'F, the point B' must lie on the line OB which bisects the angle EOF (I. 127).

etc.;

2d. If the circumscribed polygon A'B'C'D'.... is given, we have only to draw OA', OB', OC', etc., intersecting the circumference in A, B, C, etc., and then to join AB, BC, CD, etc., to obtain the inscribed polygon of the same number of sides.

4. Corollary II. If the chords AE, EB, BF, FC, etc., be drawn, a regular inscribed polygon will be formed of double the number of sides of ABCD....

If tangents are drawn at A, B, C, etc., intersecting the tangents A'B', B'C', C'D', etc., a regular circumscribed polygon will be formed of double the number of sides of A'B'C'D'....

It is evident that the area of an inscribed polygon is less than