131. Find O the center of a square by folding its diameters. Bisect the right angles at the center, then the half right angles, and so on. Then we obtain 2" equal angles around the center, and the magnitude of each of the angles is 4/2n of a right angle, n being a positive integer. Mark off equal lengths on each of the lines which radiate from the center. If the extremities of the radii are joined successively, we get regular polygons of 2" sides.

132. Let us find the perimeters and areas of these polygons. In Fig. 47 let OA and OA1 be two radii at right angles to each other. Let the radii OA2, OA3, OA4, etc., divide the right angle A1OA into 2, 4, 8.. .. parts. Draw A A1, AA2, AA3.... cutting the radii OA2, OA3, OA4. . . .at B1, B2, B3 ...respectively, at right angles. Then B1,B2, B3....are the midpoints of the respective chords. Then AA1, AA2, AA3, AA4.. . .are the sides of the inscribed polygons of 22, 23, 24___sides respectively, and OB1, OB2....... are the respective apothems.

Let OA = R, a (2") represent the side of the inscribed polygon of 2" sides, b(2n) the corresponding apothem, p(2n) its perimeter, and A (2n) its area. For the square, a(22)=R√2; p(22)=R.22 .√2; b(22)=R/2 √2; A(22)=R2.2.

Fig. 47.

For the octagon, in the two triangles AB2O and AB1A2

AB2 / B1A2 = OA / AA2. . ½AA22=R.B1A2 = R[R-b{22)]

= r(r-r/2√2) = ½R2 (2-√2), orAA2 = R = a(23)..............(1) p(23) = R.23..................(2) b(23) =OB2 = =

A(23) = ½ perimeter X apothem

= R. 22. =R2.2√2.

Similarly for the polygon of 16 sides,

p(24) = R.24.

A(24) = R2.22. and for the polygon of 32 sides, a(25) = R p(25) = R.25. b(25) = R/2

A(25) = R2.23. The general law is thus clear.

Also A(2n) = R/2. p(2n- 1).

As the number of sides is increased indefinitely the apothem evidently approaches its limit, the radius. Thus the limit of

.....is 2; for if x represent the limit, x = a quadratic which gives x = 2, or - 1; the latter value is, of course, inadmissible.

133. If perpendiculars are drawn to the radii at their extremities, we get regular polygons circumscribing the circle and also the polygons described as in the preceding article, and of the same number of sides.

Fig. 48.

In Fig. 48, let AE be a side of the inscribed polygon and FG a side of the circumscribed polygon. Then from the triangles FIE and EIO,

OE / OI = FE / EI = FG / AE ;

... FG = R AE/OI.

The values of AE and OI being known by the previous article, FG is found by substitution.

The areas of the two polygons are to one another as FG2: AE2, i. e., as R2: OI2.

134. In the preceding articles it has been shown how regular polygons can be obtained of 22, 23.. .. 2" sides. And if a polygon of m sides be given, it is easy to obtain polygons of 2n.m sides.

135. In Fig. 48, AB and CD are respectively the sides of the inscribed and circumscribed polygons of n sides. Take E the mid-point of CD and draw AE, BE. AE and BE are the sides of the inscribed polygon of 2n sides.

Fold AF, BG at right angles to A C and BD, meeting CD in F and G.

Then FG is a side of the circumscribed polygon of 2n sides.

Draw OF, OG and OE.

Let p, P be the perimeters of the inscribed and circumscribed polygons respectively of n sides, and A, B their areas, and p', P' the perimeters of the inscribed and circumscribed polygons respectively of 2n sides, and A', B' their areas.

Then p = n.AB, P=n.CD, p' = 2n.AE, P' = 2n.FG.

Because OF bisects L COE, and AB is parallel to CD,

CF / FE = CO / OE = CO / AO = CD / AB;

...CE / FE = CD+AB / AB, or 4n.CE / 4n.FE = n.CD + n.AB / n.AB

...2P / P' = P + p / p.

... p' = 2Pp / P+P

Again, from the similar triangles EIF and AHE,

EI / AH = EF / AE, or AE2 = 2AH.EF;

... 4n2.AE2 = 4n2.AB.EF, or p' = √P'p. Now,

A=2n∆A0H, B = 2n∆COE, A' = 2n∆A0E, B' = 4n∆E0E.

The triangles A OH and AOE are of the same altitude, AH,

... ∆ AOH / ∆AOE = OH/OE

Similarly,

∆AOE / ∆COE = OA / OC.

Again because AB || CD,

...∆AOH / ∆AOE = ∆AOE / ∆COE.

... A / A' = A' / B', or A' = √AB.

Now to find B'. Because the triangles COE and

FOE have the same altitude, and OF bisects the angle EOC,

∆ COE / ∆ FOE = CE / FE = OC + OE / OE and OE= OA, and OC / OA = OE / OH = ∆ AOE / ∆ AOH; ... ∆ COE / ∆ FOE = ∆ AOE + ∆ AOH / ∆ AOH.

From this equation we easily obtain 2B / B = A' + A / A;

... B' = 2AB + A + A'.

Fig. 49.

136. Given the radius R and apothem r of a regular polygon, to find the radius R and apothem r' of a regular polygon of the same perimeter but of double the number of sides.