thus find the radius and apothem of a regular polygon of eight sides, and whose perimeter is 4. Using these values in the same formulas, they will in turn make known the radius and apothem of a regular polygon of sixteen sides, having the same perimeter, 4. We give the results of these successive operations, to seven places of decimals, in the following table: From this table, we see that a circle whose circumference is 4 has for its radius 0.6366196, etc., and consequently for its diameter 1.2732392, etc. Hence, the ratio of this circumference to its diameter, or the value of, is T This value of differs less than a unit in the sixth decimal place from the true value. There are other methods, depending upon a knowledge of the higher branches of mathematics, by which this value of has been extended to more than two hundred decimal places. The value to thirty-one decimals is T = 3.1415926535897932384626433832795. The following original geometrical construction is very simple, and gives this ratio sufficiently accurate. for all practical purposes: Let ACB be the diameter of the given circle: produce it towards N; take BD and DE each equal to AB; through E draw EG perpendicular to AE, and take EF and FG each equal to AB; join AG, AF, DG, and DF. Set off on the line EN, from E, the distances EH and HK, each equal to AG; then set off, in the opposite direction, the distance KL equal to AF, and from L set off LM equal to DG; also set off MN equal to DF Then bisect EN at the point P; bisect EP at the point R, and, finally, trisect ER at the point T; then will CT be the circumference of the circle, nearly. For, by construction, we have, if we call the diameter a unit, CE=23; EL=2EH - KL=2√13 — √10; LM=√5; MN =√2. Therefore, EN 2√13- √10 + √5 + √2; CT=2}+(2 √13 — √10 + √5 + √2) = 3.1415922, etc., which is the ratio true to six decimals. For simplicity and accuracy, a better graphic method of finding this ratio can hardly be expected, or even desired. PROBLEMS, WHICH REFER TO THE THIRD AND FOURTH BOOKS. CONSTRUCTION OF PROPORTIONAL LINES. PROBLEM I. To divide a given line into any number of equal parts. To make a definite case, suppose we wish to divide AB into five equal parts. Through A draw any indefinite line AX, making an angle with AB. Lay off on this line any convenient length five times, as Ap, A P R 8 B Pq, qr, rs, sb. Join the last extremity 6 with B, and through the other points of division draw parallel to Bb, lines cutting AB in the points P, Q, R, and S (B. II., P. VI.). The straight line will thus be divided into five equal parts (B. III., T. I.). PROBLEM II. To divide a given line into parts proportional to given lines. As a definite case, suppose we wish to divide AB into three parts, which shall be to each other as the three lines P, Q, R. As in the last Problem, draw any indefinite line AX, making an angle with P. Q R A B AB. Make Ap, pq, qb respectively equal to P, Q, and R. Draw Bb, and parallel to it draw pP', qQ', and the line AB will be divided at P' and Q' as required (B. III., T. III.) Scholium. As a particular case, suppose we wish to divide a straight line AB into two parts proportional to the lines M and N. The straight line AB will thus be divided at D in the ratio required. For, the two triangles DAC, DBC', are evidently similar, and give AD: DB::AC BC'::M:N. If, instead of taking BC', equal to N, in the direction of BY, opposite to AX, we had taken BC", equal to N, in the same direction with AX, then the line CC" would have met AB produced at D', which is called the conjugate point of the point D (B. III., T. XII., S.). PROBLEM III. To find a fourth proportional to three given lines M, N, P. Form any angle, as XAY, and take on AX, AB = M, BC = N, and on AY, AD=P; then draw BD, and through the point C draw CE parallel to BD. The line DE will be the fourth proportional required. For, we have (B. III., T. III.), Y E D A -X B C M N P AB: BC:: AD: DE, or M:N::P: DE. Cor. I. If P were equal to N, the above proportion would become M:N::N: DE. That is, DE would in this case be a third proportional to the two lines M and N. Cor. II. We deduce immediately the following problem: A point O being given within an angle YAX, to draw through O a straight line DOE, such that the segments DO, OE may be to each other in the ratio of M to N. and draw EOD, which will be the line required. For, since OB is parallel to AD, we have OD: OE::AB:BE::M: N. C X B E When M = N, it is sufficient to take on AX, BE = AB, and then to draw EOD. PROBLEM IV. To find a mean proportional between two given lines M AB: BD::BD: BC, or M: BD:: BD: N. PROBLEM V. To divide a given line into mean and extreme ratio. We have already noticed this kind of division' (T. III., S. I.). And we have also actually solved this problem algebraically (T. XI., S. I.). |