(2) Divide the line AB in C, so that the rectangle AB, BC, may be equal to the square on AC.

(3) Describe the circle BDE with center A and radius AB. (4) Place the line BD in that circle, equal to the line AC. (5) Join the points A, D.

Why should either of these operations be performed rather than any others? And what will enable us to forsee that the result of them will be such a triangle as was required? The demonstration affixed to it by Euclid does undoubtedly prove that these operations must, in conjunction, produce such a triangle: but we are furnished in the Elements with no obvious reason for the adoption of these steps, unless we suppose them accidental. To suppose that all the constructions, even the simpler ones, are the result of accident only, would be supposing more than could be shewn to be admissible. No construction of the problem could have been devised without a previous knowledge of some of the properties of the figure. In fact, in directing the figure to be constructed, we assume the possibility of its existence; and we study the properties of such a figure on the hypothesis of its actual existence. It is this study of the properties of the figure that constitutes the Analysis of the problem.

Let then the existence of a triangle BAD be admitted, which has each of the angles ABD, ADB double of the angle BAD, in order to ascertain any properties it may possess which would assist in the construction of such a triangle.

Then, since the angle ADB is double of BAD, if we draw a line DC to bisect ADB and meet AB in C, the angle ADC will be equal to CAD; and hence (Euc. 1. 6.) the sides AC, CD are equal to one another.

Again, since we have three points A, C, D, not in the same straight line, let us examine the effect of describing a circle through them: that is, describe the circle ACD about the triangle ACD. (Euc. IV. 5.)

Then, since the angle ADB has been bisected by DC, and since ADB is double of DAB, the angle CDB is equal to the angle DAC in the alternate segment of the circle; the line BD therefore coincides with a tangent to the circle at D. (Converse of Euc. 111. 32.)

Whence it follows, that the rectangle contained by AB, BC, is equal to the square on BD. (Euc. III. 36.)

But the angle BCD is equal to the two interior opposite angles CAD, CDA; or since these are equal to each another, BCD is the double of CAD, that is, of BAD. And since ABD is also double of BAD, by the conditions of the triangle, the angles BCD, CBD are equal, and BD is equal to DC, that is, to AC.

It has been proved that the rectangle AB, BC, is equal to the square on BD; and hence the point C in AB, found by the intersection of the bisecting line DC, is such, that the rectangle AB, BC is equal to the square on AC. (Euc. 11. 11.)

Finally, since the triangle ABD is isosceles, having each of the angles ABD, ADB double of the same angle, the sides AB, AD are equal, and hence the points B, D, are in the circumference of the circle described about A with the radius AB. And since the magnitude of the triangle is not specified, the line AB may be of any length whatever.

From this "Analysis of the problem," which obviously is nothing more than an examination of the properties of such a figure supposed to exist already, it will be at once apparent, why those steps which are prescribed by Euclid for its construction, were adopted.

The line AB is taken of any length, because the problem does not prescribe any specific magnitude to any of the sides of the triangle.

The circle BDE is described about A with the distance AB, because the triangle is to be isosceles, having AB for one side, and therefore the other extremity of the base is in the circumference of that circle.

The line AB is divided in C, so that the rectangle AB, BC shall be equal to the square on AC, because the base of the triangle must be equal to the segment AC.

And the line AD is drawn, because it completes the triangle, two of whose sides, AB, BD are already drawn.

Whenever we have reduced the construction to depend upon problems which have been already constructed, our analysis may be terminated; as was the case where, in the preceding example, we arrived at the division of the line AB in C; this problem having been already constructed as the eleventh of the second book.

Prop. xvi. The arc subtending a side of the quindecagon, may be found by placing in the circle from the same point, two lines respectively equal to the sides of the regular hexagon and pentagon.

The centers of the inscribed and circumscribed circles of any regular polygon are coincident.

Besides the circunscription and inscription of triangles and regular polygons about and in circles, some very important problems are solved in the constructions respecting the division of the circumferences of circles into equal parts.

By inscribing an equilateral triangle, a square, a pentagon, a hexagon, &c. in a circle, the circumference is divided into three, four, five, six, &c. equal parts. In Prop. 26, Book III, it has been shewn that equal angles at the centers of equal circles, and therefore at the center of the same circle, subtend equal arcs; by bisecting the angles at the center, the arcs which are subtended by them are also bisected, and hence, a sixth, eighth, tenth, twelfth, &c. part of the circumference of a circle may be found.

If the right angle be considered as divided into 90 degrees, each degree into 60 minutes, and each minute into 60 seconds, and so on, according to the sexagesimal division of a degree; by the aid of the first corollary to Prop. 32, Book 1, may be found the numerical magnitude of an interior angle of any regular polygon whatever.

Let 0 denote the magnitude of one of the interior angles of a regular polygon of n sides,

then no is the sum of all the interior angles.

But all the interior angles of any rectilineal figure together with four right angles, are equal to twice as many right angles as the figure has sides, that is, if π be assumed to designate two right angles,

the magnitude of an interior angle of a regular polygon of n sides. By taking n = 3, 4, 5, 6, &c. may be found the magnitude in terms of two right angles, of an interior angle of any regular polygon whatever.

Pythagoras was the first, as Proclus informs us in his commentary, who discovered that a multiple of the angles of three regular figures only, namely, the trigon, the square, and the hexagon, can fill up space round a point in a plane.

It has been shewn that the interior angle of any regular polygon of n

sides in terms of two right angles, is expressed by the equation

Let s denote the magnitude of the interior angle of a regular figure of three sides, in which case, n = 3.

that is, six angles, each equal to the interior angle of an equilateral triangle, are equal to four right angles, and therefore six equilateral triangles may be placed so as completely to fill up the space round the point at which they meet in a plane.

In a similar way, it may be shewn that four squares and three hexagons may be placed so as completely to fill up the space round a point.

Also it will appear from the results deduced, that no other regular figures besides these three, can be made to fill up the space round a point; for any multiple of the interior angles of any other regular polygon, will be found to be in excess above, or in defect from four right angles.

The equilateral triangle or trigon, the square or tetragon, the pentagon, and the hexagon, were the only regular polygons known to the Greeks, capable of being inscribed in circles, besides those which may be derived from them.

M. Gauss in his Disquisitiones Arithmeticæ, has extended the number by shewing that in general, a regular polygon of 2" + 1 sides is capable of being inscribed in a circle by means of straight lines and circles, in those cases in which 2" + 1 is a prime number.

The case in which n = 4, in 2" + 1, was proposed by Mr. Lowry of the Royal Military College, to be answered in the seventeenth number of Leybourn's Mathematical Repository, in the following form:

Required a geometrical demonstration of the following method of constructing a regular polygon of seventeen sides in a circle.

Draw the radius CO at right angles to the diameter AB; on OC and OB, take OQ equal to the half, and OD equal to the eighth part of the radius; make DE and DF each equal to DQ, and EG and FH respectively equal to EQ and FQ; take OK a mean proportional between OH and OQ, and through K, draw KM parallel to AB, meeting the semicircle described on OG in M, draw MN parallel to OC cutting the given circle in N, the arc AN is the seventeenth part of the whole circumference.

A demonstration of the truth of this construction has been given by Mr. Lowry himself, and will be found in the fourth volume of Leybourn's Repository. The demonstration including the two lemmas occupies more than eight pages, and is by no means of an elementary character.

QUESTIONS ON BOOK IV.

1. WHAT is the general object of the Fourth Book of Euclid? 2. What consideration renders necessary the first proposition of the Fourth Book of Euclid?

3. When is a circle said to be inscribed within. and circumscribed about a rectilineal figure?

4. When is one rectilineal figure said to be inscribed in, and circum scribed about another rectilineal figure?

5. Modify the construction of Euc. rv. 4, so that the circle may touch one side of the triangle and the other two sides produced.

6. The sides of a triangle are 5, 6, 7 units respectively, find the radii of the inscribed and circumscribed circle.

7. Give the constructions by which the centers of circles described about, and inscribed in triangles are found. In what triangles will they coincide?

8. How is it shown that the radius of the circle inscribed in an equilateral triangle is half the radius described about the same triangle? 9. The equilateral triangle inscribed in a circle is one-fourth of the equilateral triangle circumscribed about the same circle.

10. What relation subsists between the square inscribed in, and the square circumscribed about the same circle?

11. Enunciate Euc. 11. 22: and extend this property to any inscribed polygon having an even number of sides.

m

12. Trisect a quadrantal arc of a circle, and show that every arc which is an th part of a quadrantal arc may be trisected geometrically: m and n being whole numbers:

2n

13. If one side of a quadrilateral figure inscribed in a circle be produced, the exterior angle is equal to the interior and opposite angle of the figure. Is this property true of any inscribed polygon having an even number of sides?

14. In what parallelograms can circles be inscribed?

15. Give the analysis and synthesis of the problem: to describe an isosceles triangle, having each of the angles at the base double of the third angle?

16. Shew that in the figure Euc. Iv. 10, there are two triangles possessing the required property.

17. How is it made to appear that the line BD is the side of a regular decagon inscribed in the larger circle, and the side of a regular pentagon inscribed in the smaller circle? fig. Euc. iv. 10.

18. In the construction of Euc. Iv. 3, Euclid has omitted to shew that the tangents drawn through the points A and B will meet in some point M. How may this be shewn?

19. Shew that if the points of intersection of the circles in Euclid's figure, Book IV. Prop. 10, be joined with the vertex of the triangle and with each other, another triangle will be formed equiangular and equal to the former.

20. Divide a right angle into five equal parts. How may an isosceles triangle be described upon a given base, having each angle at the base one-third of the angle at the vertex?

21. What regular figures may be inscribed in a circle by the help of Euc. IV. 10?

22. What is Euclid's definition of a regular pentagon? Would the stellated figure, which is formed by joining the alternate angles of a regular pentagon, as described in the Fourth Book, satisfy this definition?

23. Shew that each of the interior angles of a regular pentagon inscribed in a circle, is equal to three-fifths of two right angles.

24. If two sides not adjacent, of a regular pentagon, be produced to meet: what is the magnitude of the angle contained at the point where they meet?

25. Is there any method more direct than Euclid's for inscribing a regular pentagon in a circle?

26. In what sense is a regular hexagon also a parallelogram? Would the same observation apply to all regular figures with an even number of sides?

27. Why has Euclid not shewn how to inscribe an equilateral triangle in a circle, before he requires the use of it in Prop. 16, Book IV.?

28. An equilateral triangle is inscribed in a circle by joining the first, third, and fifth angles of the inscribed hexagon.

29. If the sides of a hexagon be produced to meet, the angles formed by these lines will be equal to four right angles.

30. Shew that the area of an equilateral triangle inscribed in a circle is one-half of a regular hexagon inscribed in the same circle.

31. If a side of an equilateral triangle be six inches: what is the radius of the inscribed circle?

32. Find the area of a regular hexagon inscribed in a circle whose diameter is twelve inches. What is the difference between the inscribed and the circumscribed hexagon ?

33. Which is the greater, the difference between the side of the square and the side of the regular hexagon inscribed in a circle whose radius is unity; or the difference between the side of the equilateral triangle and the side of the regular pentagon inscribed in the same circle?

54. The regular hexagon inscribed in a circle, is three-fourths of the regular circumscribed hexagon.

35. Are the interior angles of an octagon equal to twelve right angles? 36. What figure is formed by the production of the alternate sides of a regular octagon ?

37. How many square inches are in the area of a regular octagon whose side is eight inches?

38. If an irregular octagon be capable of having a circle described about it, shew that the sums of the angles taken alternately are equal.

39. Find an algebraical formula for the number of degrees contained by an interior angle of a regular polygon of n sides.

40. What are the three regular figures which can be used in paving a plane area? Shew that no other regular figures but these will fill up the space round a point in a plane.

41. Into what number of equal parts may a right angle be divided geometrically? What connection has the solution of this problem with the possibility of inscribing regular figures in circles?

42. Assuming the demonstrations in Euc. IV, shew that any equilateral figure of 3.2", 4.2", 5.2", or 15.2" sides may be inscribed in a circle, when n is any of the numbers, 0, 1, 2, 3, &c.

43. With a pair of compasses only, shew how to divide the circumference of a given circle into twenty-four equal parts.

44. Shew that if any polygon inscribed in a circle be equilateral, it must also be equiangular. Is the converse true?

45. Shew that if the circumference of a circle pass through three angular points of a regular polygon, it will pass through all of them. 46. Similar polygons are always equiangular: is the converse of this proposition true?

47. What are the limits to the Geometrical inscription of regular figures in circles? What does Geometrical mean when used in this way? 48. What is the difficulty of inscribing geometrically an equilateral and equiangular undecagon in a circle? Why is the solution of this problem said to be beyond the limits of plane geometry? Why is it so difficult to prove that the geometrical solution of such problems is impossible?