Section 2.Partition of the Perigon.

Problem 1. To bisect a perigon.

Construction and Proof. This problem and the corollary are merely special cases under I, pr. 1 and cor.

COROLLARY. A perigon can be divided into 2" equal angles.

4. Then the perigon is trisected by OB, OC, OD.

A

I, th. 19, cor. 8

Ax. (?)

Ax. (?)

Const. 3

.. COB 240°, by subtracting from

COROLLARY. A perigon can be divided into 3.2" equal angles.

For if n = 0, 3.2′′ = 3·1 = 3, the above problem. If n = 1, 3·2n = 6, and by bisecting BOC, COD, DOB, the perigon is divided into 6 equal angles. Similarly, by bisecting again, the perigon is divided into 3-22=12 equal angles, and so on.

Problem 3. To divide a perigon into fiv

Given the perigon with vertex O.

to divide it into five equal

Construction. 1. Draw OA, and divide

it at P so that OP OA=PA2.

II, pr. 6, or p. 168

2. Draw MY, the

OP.

bisector of

I, pr. 4

3. With center P and radius PA descri

MY in B.

3. Also, OA+OB2 AB2+2 OM OA

=

7. ..ZOBA=0=/ BPO.

... OB2

10. O is of a st. Z, or of a pe

COROLLARY. A perigon can be divided into Explain.

EXERCISES. 458. In the above figure, let OP = x,

that x= (√5-1). (Omit exs. 458, 459 if the st quadratic equations.)

459. In the same figure, if OP x and OA = (3- √5).

Problem 4. To divide a perigon into fifteen equal angles.

COROLLARY. A perigon can be divided into 15·2′′ equal angles. Explain.

NOTE. That a perigon could be divided into 2", 3.2", 5.2", 15.2" equal angles, was known as early as Euclid's time. By the use of the compasses and straight edge no other partitions were deemed possible. In 1796 Gauss found, and published the fact in 1801, that a perigon could be divided into 17, and hence into 17·2′′ equal angles; furthermore, that it could be divided into 2m+1 equal angles if 2m+1 was a prime number; and, in general, that it could be divided into a number of equal angles represented by the product of different prime numbers of the form of 2m+1. Hence it follows that a perigon can be divided into a number of equal angles represented by the product of 2′′ and one or more different prime numbers of the form 2m+1. It is shown in the Theory of Numbers that if 2m+1 is prime, m must equal 2o; hence the general form for the prime numbers mentioned is 222 + 1. Gauss's proof is only semigeometric, and is not adapted to elementary geometry.

EXERCISES. 460. Including the divisions of a perigon suggested by Gauss, there are 25 possible divisions below 100. What are they?

461. As in ex. 460, there are 13 possible divisions between 100 and 300. What are they?

462. The line joining any point, on a circumference circumscribing an equilateral triangle, to the farthest vertex of that triangle, equals the sum of the lines joining that point to the other two vertices. (In th. 9, step 9,

f=c=d, if ACD is an equilateral triangle, and B is the point.)

463. If A and C are fixed points, and D is any variable point, all on a circumference, and if B is the mid-point of arc AC, then the ratio of DA+ DC to DB is constant.

Section 3.- Regular Poly

Problem 5. To inscribe in a circle a having a given number of sides.

Given a circle with center O and radius OA.

Required to inscribe in the circle a regular n-gon.

Construction. 1. Divide the perigon O into n equal parts, n being limited as in prs. 1-4 and cors., as AOB, BOC, COD,

B, C, D,..... lying on the circ 2. Draw AB, BC, CD.....

3. Then ABCD ..... is an inscribed re

A AOBA BOCA COD=

and AB BCCD=

DCB = CBA = =

=

= .....

each stands on (n-2) arcs equ

4... ABCD is an inscribed regula

COROLLARIES. 1. The side of an inscribed

equals the radius of the circle.

2. An inscribed equilateral polygon is regul For by step 3 of the above proof it is also equiang

EXERCISE. 464. If r is the radius of the circle, an inscribed equilateral triangle, then s = r =r√3.

Problem 6. To circumscribe about a circle a regular polygon having a given number of sides.

.....

A

DW

2. BisectWOX, XOY,..... by radii to A, B, .. I, pr. 1 3. From A, B, C,..... draw tangents to meet OW at D, OX at E, .....

4. Then DEFG..... is the required polygon.

Proof. 1. DE LOA, EF LOB,.....

III, pr. 3

III, th. 9, cor. 2

2...A OAEA OBE, and AE BE, QE=OE.

=

I, th. 2

3... the tangents from A and B meet OX at the same

6. Also, GFE FED, each being the supplement

of an equal to

WOX. (Name the .)

I, th. 21, cor.

Def.

7... DEFG... is a circumscribed regular polygon.

COROLLARIES. 1. The side of a regular hexagon circumscribed about a circle of diameter 1, is 1/ √3, or 1 √3.

For it is (as in pr. 5, cor. 1) the side of an equilateral ▲ whose altitude is. This is easily shown to be 1/√3. (Show it.)

2. A circumscribed equiangular polygon is regular.

Prove that any two adjacent sides are equal.