that extent form the square a moh, which will be the square required. With dh you may form a square that will contain the given one aefd four times. The line d 1 gives a square five times larger than the original square.

PROBLEM 67. To reduce a mapì, ìd, i̟th, th, &c. of the original, fig. 5, plate 8.

1. Divide the given plan into squares by problem 65. 2. Draw a line, on which set off from A to B one side of one of these squares. 3. Divide this line into two equal parts at F, and on Fas a centre, with FA or F B, describe the semicircle AH B.

5.

4. To obtain the given square, at F erect the perpendicular FH, and draw the right line AH, which will be the side of the required square. For rd, divide A B into three parts, take one of these parts, set it off from A to C, at C raise the perpendicular C I, through I draw AI, and it will be the side of the required square. 6. For 4th, divide A B into four equal parts, set off one of these from B to E, at E make EG perpendicular to A B, join B G, and it will be the side of a square 4th of the given

one.

PROBLEM 68. To make a map or plan in proportion to a given one, ex. gr. as three to five, fig. 6, plate 8.

The original plan being divided into squares, 1. draw A M equal to the side of one of these squares. 2. Divide A M into five equal parts. 3. At the third division raise the perpendicular CD, and draw AD, which will be the side of the required square.

PROBLEM 69. To reduce figures by the angle of reduction. Let ab be the given side on which it is required to construct a figure similar to ACDEB, fig. 1, 2, 3, plate 8.

1. Form an angle L M N at pleasure, and set off the side A B from M to I. 2. From I, with a b, cut ML in K. 3. Draw the line I K, and several lines parallel to, and on both sides of it. The angle

LMN is called the angle of proportion or reduction. 4. Draw the diagonal lines BC, AD, AE, B D. 5. Take the distance BC, and set it off from M towards L on ML. 6. Measure its corresponding line KI. 7. From b describe the arc n o. 8. Now take AC, set it off on M L, and find its correspondent line fg. 9. From a, with the radius fg, cut the former are no in c, and thus proceed till you have completed the figure.

PROBLEM 70. To enlarge a figure by the angle of reduction. Let a bede be the given figure, and AB the given side, fig. 3, 1, and 4, plate 8.

1. Form, as in the preceding problem, the angle LMN, by setting off a b from M to H, and from Hwith AB, cutting ML in I. 2. Draw H I, and parallels to, and on both sides of it. 3. Take the diagonal bc, set it off from M towards N, and take off its corresponding line qr. 4. With qr as a radius on B describe the arc m n. 5. Take the same correspondent line to a c, and on A cut mn in C, and so on for the other sides.

CURIOUS PROBLEMS, ON THE DIVISION OF LINES AND CIRCLES.

PROBLEM 71. To cut off from any given arc of a circle a third, a fifth, a seventh, &c. odd parts, and thence to divide that arc into any number of equal parts, fig. 7, plate 8.

Example 1. To divide the arc A KB into three equal parts, CA being the radius, and C the centre

of the arc.

Bisect AB in K, draw the two radii C K, CB, and the chord A B; produce A B at pleasure, and make BL equal A B; bisect A C at G; then a rule on G and L will cut CB in E, and B E will be id, and C Eds of the radius CB; on CB with CE describe the arc Ee D; lastly, set off the extent Ee

or De from B to a, and from a to b, and the arc AKB will be divided into three equal parts.

Corollary. Hence having a sextant, quadrant, &c. accurately divided, the chord of any arc set off upon any other are of that radius will cut off an arc 14 similar to the first, and containing the same number of degrees.

Also id, 4th, th, &c. of a larger chord will constantly cut similar arcs on a circle whose radius isd, 4th, th, &c. of the radius of the first arc.

Example 2. Let it be required to divide the arc AKB into five equal parts, or to find the 4th part of the arc A B.

Having bisected the given arc A B in K, and drawn the three radii CA, CK, CB, and having found the fifth part of IB of the radius CB, with radius CI describe the arc In M, which will be bisected in n, by the line CK; then take the extent In, or its equal M n, and set it off twice from A to B; that is, first, from A to d, and from d to o, and o B will be 4th of the arc AB. Again, set off the same extent from B to m, and from m to c, and the arc AB will be accurately divided into five equal parts.

Example 3. To divide the given arc AB into seven equal parts. AB being bisected as before, and the radii CA, CK, CB, drawn, find by problem 9 the seventh part PB of the radius CB, and with the radius CP, describe the arc PrN; then set off the extent Pr twice from A to 3, and from 3 to 6, and 6 B will be the seventh part of the given arc AB; the compasses being kept to the same opening Pr, set it from B to 4, from 4 to 1; then the extent A 1 will bisect 1, 3 into 2, and 4, 6 into 5; and thus divide the given arc into seven equal parts.

PROBLEM 72. To inscribe a regular heptagon in a circle, fig. 8, plate 8.

In fig. 8, let the arc B D be 4th part of the given

circle, and AB the radius of the circle. Divide A B into eight equal parts, then on centre A, with radius AC, describe the arc CE, bisect the arc BD in a, and set off this arc B a from C to b, and from b to c; then through A and c draw A ce, cutting BD in e, and Be will be 4th part of the given circle.

Corollary 1. Hence we have a method of finding the seventh part of any given angle; for, if from the extremities of the given arc, radii be drawn to the centre, and one of these be divided into eight equal parts, and seven of these parts be taken, and another arc described therewith, the greater arc will be to the lesser, as 8 to 7, and so of any other proportion.

Corollary 2. If an arc be described with the radius, it will be equal to the 4th part of a circle whose radius is AB, and to the seventh part of a circle' whose radius is AC, and to the sixth part of a circle whose radius is A L, &c.

Corollary 3. Hence also a pentagon may be derived from a hexagon, fig. 9, plate 8. Let the given circle be ABCDEF, in which a pentagon is to be inscribed; with the radius A C set off A F equal th of the circle, divide A C into six equal parts; then CG will be five of these parts; with radius C G describe the arc G H, bisect A F in q, and make Gp and p H each equal to A q; then through C and H draw the semi-diameter C w, cutting the given circle in w, join A w, and it will be one side of the required pentagon.

Corollary 4. Hence as radius divides a circle into six equal parts, each equal 60 degrees, twice radius gives 120 degrees, or the third part of the circumference.

Once radius gives 60 degrees, and that are bisected gives 30 degrees, which, added to 60, divides the circumference into four equal parts; whence we divide it into two, three, four, five, six equal parts; the preceding corollary divides it into five equal parts, the arc of a quadrant bisected divides it into eight

equal parts. By problem 71 we obtain the seventh part of a circle, and by this method divide it into any number of equal parts, even a prime number; for the odd unit may be cut off by the preceding problem, and the remaining part be subdivided by continual bisection, till another prime number arises to be cut off in the same manner.

PROBLEM 73. To divide a given right line, or an arc of a circle, into any number of equal parts by the help of a pair of beam, or other compasses, the distance of whose points shall not be nearer to each other than the given line, fig. 11, plate 8.

From this problem, published by Clavius, the Jesuit, in 1611, in a treatise on the construction of a dialling instrument, it is presumed that he was the original inventor of that species of division, called Nonius, and which by many modern mathematicians has been called the scale of Vernier.

Let A B be the given line, or circular arch, to be divided into a number of equal parts. Produce them at pleasure; then take the extent A B, and set it off on the prolonged line, as many times as the given line is to be divided into smaller parts, BC, CD, DE, EF, FG. Then divide the whole line A G into as many equal parts as are required in A B, as GH, HI, ÍK, KL, LA, each of which contains the given line, and one of those parts into which the given line is to be divided. For A G is to A Las AF to AB; in other words, AL is contained five times in A G, as A Bin AF; therefore, since A G, contains A F, and 4th of A B; therefore B L is the th of A B. Then as G H contains A B plus FH, which is 4th of AB, EI will be ths of AB, D K ths, CL ths. Therefore, if we set off the interval GH from F and H, we obtain two parts at L and I, set off from the points near E, gives three parts between D K, from the four points at D and K, gives four parts at C L, and the next setting off one more of these parts; so that lastly, the extent G H set off