sides in terms of two right angles, is expressed by the equation = TT, Let 0, denote the magnitude of the interior angle of a regular figure of three sides, in which case, n = 3. 3 – 2 Then og = one third of two right angles, 3 3 .. 303 : and 60, = 27, 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 27 + 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 of and OQ, and through K, draw KM parallel to AB, meeting the semicircle described on Oã 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 ? K m 4. When is one rectilineal figure said to be inscribed in, and circumscribed about another rectilineal figure ? 5. Modify the construction of Euc. IV. 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. III. 22: and extend this property to any inscribed polygon having an even number of sides. 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: 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 ? а. 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. All 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 ? PROPOSITION I. THEOREM. If an equilateral triangle be inscribed in a circle, the square of the side of the triangle is triple of the square of the radius, or of the side of the regular hexagon inscribed in the same circle. Let ABD be an equilateral triangle inscribed in the circle ABD, of which the center is C. А E B Join BC, and produce BC to meet the circumference in E, also join AE. And because ABD is an equilateral triangle inscribed in the circle; therefore AED is one-third of the whole circumference, and therefore AE is one-sixth of the circumference, and consequently, the straight line AE is the side of a regular hexagon (IV. 15.), and is equal to EC. And because BE is double of EC or AE, therefore the square on BE is quadruple of the square on AE, but the square on BE is equal to the squares on AB, AE; therefore the squares on AB, AĒ are quadruple of the square on A E, and taking from these equals the square on AE, therefore the square on AB is triple of the square on AE PROPOSITION II. PROBLEM. To describe a circle which shall touch a straight line given in position, and pass through two given points. Analysis. Let AB be the given straight line, and C, D the two given points. Suppose the circle required which passes through the points C, D to touch the line AB in the point E. Join C, D, and produce DC to meet AB in F, and let the circle be described having the center L, join also LE, and draw LH perpendicular to CD. Then CD is bisected in H, and LE is perpendicular to AB. Also, since from the point F without the circle, are drawn two straight lines, one of which FE touches the circle, and the other FDC cuts it; the rectangle contained by FC, FD, is equal to the square of FE. (III. 36.) Synthesis. Join C, D, and produce CD to meet AB in F, take the point E in FB, such that the square on FE, shall be equal to the rectangle FD, FC. At E draw EĞ perpendicular to FB, then EG passes through the center, (111. 19.) consequently L, the point of intersection of these two lines, is the center of the circle. It is also manifest, that another circle may be described passing through C, D, and touching the line AB on the other side of the point F; and this circle will be equal to, greater than, or less than the other circle, according as the angle CFB is equal to, greater than, or less than the angle CFA. a PROPOSITION III. PROBLEM. Inscribe a circle in a given sector of a circle. Analysis. Let CAB be the given sector, and let the required circle whose center is 0, touch the radii in P, Q, and the arc of the sector in D. с Join OP, OQ, these lines are equal to one another. Join also CO. Then in the triangles CPO, CQO, the two sides PC, CO, are equal to QC, CO, and the base OP is equal to the base OQ; therefore the angle PCO is equal to the angle QCO; and the angle ACB is bisected by CO: and CA, CB be produced to meet it in E, F: tion of a circle in a triangle. (IV. 4.) PROPOSITION IV. PROBLEM. ABCD is a rectangular parallelogram. Required to draw EG, FG parallel to AD, DC, so that the rectangle EF may be equal to the figure EMD, and EB equal to FD. Analysis. Let EG, FG be drawn, as required, bisecting the rectangle ABCD. |