Problem 2

343. Inscribe a regular hexagon in a given circle.

Given a circle whose center is K.

Required to inscribe a regular hexagon in the circle.

Construction. Draw any radius KA. Then with A as the center and AK as the radius, describe an arc cutting the circle at B. With B as the center and KA as the radius, describe an arc cutting the circle at C. In like manner obtain points D, E, and F. Draw the chords AB, BC, CD, DE, EF, and FA. They will form the required regular hexagon.

Proof. Draw the radius KB.

Then

Hence

and

AABK is equilateral.
<K = 60°,

arc AB = 60°, or one sixth of the circumference.

Why?

Why?

Why?

Hence arc ABCDEF is five sixths of the circle, and arc FA is one sixth.

In like manner ZE, ZF, etc. are each equal to 120°,

and

in the circle.

the hexagon is equiangular.

$ 178

Why?

Therefore the polygon ABCDEF is a regular hexagon inscribed

$ 79

EXERCISES

In Exs. 11-18 the result of any one may be used, if needed, in any one following.

R

2

11. Inscribe an equilateral triangle in a given circle.

12. Inscribe a regular dodecagon in a given circle.

13. Circumscribe a regular hexagon about a given circle.

14. Circumscribe an equilateral triangle about a circle.

15. Show that the apothem of a regular inscribed hexagon is √3 and that the area is

16. Show that the side

3 R2

2

R

of a regular inscribed dodecagon is

R√2 −√3, the apothem √2+√3, and the area 3 R2.

2

17. Show that the side, the apothem, and the area of a regular inscribed dodecagon in a circle whose radius is 20 inches are respectively 10.35+ inches, 19.31+ inches, and 1200 square inches.

19. The radius of a circle is 18. Show that the area of its regular circumscribed hexagon is 1122.36 + square units.

20. Show that the side, the apothem, and the area of an inR 3 R2√3 scribed equilateral triangle are respectively R√3, and 2?

4

21. Show that the side, the apothem, and the area of a circumscribed equilateral triangle are respectively 2 R √3, R, and 3 R2 √3.

344. Mean and extreme ratio. A line is divided by a point in mean and extreme ratio when one segment is a mean proportional between the whole line and the other segment.

There are two cases: (1) when the point K is on the given line AB (Fig. 1); (2) when the point R is on the given line BA produced (Fig. 2):

As case (2) is not used in elementary geometry, it will be given no further consideration here.

Problem 3 (Computation)

345. Express the numerical values of the two segments of a line of length a divided internally in mean and extreme ratio by a point on the line.

Given the line AB of length a, and the point K dividing the line in extreme and mean ratio.

Required to express the values of the segments AK and KB in terms of a.

Solution. Let K be the point which divides AB internally in mean and extreme ratio. Then, by the definition of § 344,

Since (8) gives a negative value of x, it is neither AK nor BK of the figure. Therefore (7) gives the length of AK.

a

(7)

(8)

EXERCISE 22.

Show that the segments of a line 12' long divided internally in mean and extreme ratio are 7.42'- and 4.57' respectively.

NOTE. The problem of § 345 was of great interest to the early Greek geometers. Plato (429-347 B.C.) was the first to discuss and solve the problem. Later Eudoxus (408-355 B.C.), who is usually ranked as third among the Greek mathematicians, added several other theorems concerning it. Euclid in the thirteenth book of his "Elements" has five theorems on the "golden section." This name was used by the Greeks to designate what now is usually called mean and extreme ratio. The Greeks knew little of modern algebra and nothing of the Hindu notation. Hence their proofs are very much more difficult in appearance than the proof given of Problem 3. The actual geometrical construction, however, follows in § 346.

The numerous applications of the "golden section" in architecture and other forms of art indicate that this division of a line possesses æsthetic as well as mathematical significance.

Required to locate a point K on AB such that AB: AK=AK:KB. Construction. Bisect AB at L. Construct BMI to AB at B. Lay off BG equal to BL. Construct a circle with G as the center and BG as the radius. Draw AG, cutting the circle in F and H. Lay off AK on AB equal to AF.

347. Decagon. A decagon is a polygon having ten sides.