Discussion 3.-In like manner, regarding Figure 2, as BD must always be of the length of the given applied line, and DE equal to the radius BH, it will be seen that, as the applied line is taken longer and longer, D recedes from B, AD increases, E and E' approach each other, till the angles ADE and ADE' vanish, E and E' meet at E", when DE" DE DE' BH, and then the applied line will be the longest possible, or a maximum. To obtain this maximum line by construction, with centre A (Fig. 3) and radius equal to the sum of the two radii AG and BH, describe an arc, cutting BC in D', further from B than the foot of the perpendicular AO let fall from A on BC. Draw AE'D', and draw E'F' parallel to BC. Join BF; then will BD'E'F' be a parallelogram, and E'F' will evidently be equal and parallel to BD', and be the required maximum line. To determine its length by calculation, in triangle ABD', Case 2, find BD' — E'F". Discussion 4.-Through A (Fig. 3), parallel to BC, draw a line meeting BF and BF", produced, in I and L respectively; then, since AD AD', and AE= AE', EE' is parallel to BC. But EF is parallel to BC. Hence E'EF is a straight line. Again, since BL AD' and BF" — D'E', we have F'L: AE'. In like manner BI - AD and BF=ED. Therefore FI — AE AE' FL. Hence FF is parallel to IL or BC. But EF is parallel to BC; wherefore the three lines E'E, EF, and FF" are in one and the same straight line, parallel to BC. Therefore, the shortest line possible (EF), and the longest line possible (E'F"), that can be applied between the circles, parallel to a line given in length and position, are in the same straight line, the minimum (EF) being part of the maximum (E'F"). Discussion 5.-In case of the minimum and maximum lines EF and E'F' (Fig. 3), the angle EAG — HBF, and arc EG is similar to arc HF. Also, the angle E'AG — HBF", and arc E'G is similar to arc HF". Also, ADBI, AEFI, AD'BL, AE'F'L are parallel ograms. Discussion 6.—The maximum angle which the line BC, given in position, can make with the line AB, is that made by the internal common tangent to the two given circles, as EPF, E'PF' (Fig. 4). For it is evident that a line drawn from any point in the circle BH, making with AB an angle greater than BPF or BPF", will not meet the circle AG; and that a line drawn from any point in the circle AG, making with AB an angle greater than APE or APE', will not meet the circle BH. FIG. 4. E P H E' I I' B These two lines EF and E'F' are equal; they intersect each other in the same point P on AB; and are the only lines which can be applied between the circles, making with AB an angle as great as APE or APE'. To find these lines by construction.—On AB as a diameter describe a circle, in which apply AO and AO' equal to the sum of the radii AG and BH. Draw BOC, and join BO', parallel to which lines, respectively, draw EF and E'F', and join BF and BF"; then (Problem XIII. of "The Circle") EF and E'F' will be tangents to both circles. To find the lengths of these common tangents by calculation.-We have EF: E'F' BO: =√ AB2 AB2 — AO2. Corollary. If BF and BF be produced to meet the circle described on AB in I' and I, then Fl' AE = AE' F'I, and EF is parallel to AI' as well as to BO, also E'F' is parallel to AI and to BO'. Discussion 7.—The minimum angle which the line BC, given in FIG. 5 E C G DH E! F B position, can make with AB is 0°. In this case BC (Fig. 5) will coincide with BA, and, making BD the length of the given line, and applying DE, DE' each equal to BH, and drawing EF and E'F' each parallel to BD, we have the construction just as in Figure 1. Calculation. In triangle AED, Case 4, find angle ADE=DEF ADE'; then EDB 180° ADE E'DB. DE'F PROBLEM XXXII.—To produce a line of a given length so that the rectangle of the whole line thus produced, and the part produced, shall be equal to the square of a given line. (See IV. Prob. 4.*) E Given the length of ED, to produce it so that EA × AD=AB2. A B Analysis. It is evident that AB is tangent to the circle whose radius CB is half the given line ED, to be produced so that EA × AD AB2. Whence this the given line. Through A and C Construction.-With a radius equal to half the length of the given line to be produced, describe a circle, as BDE. Draw any radius, as CB, perpendicular to which draw the tangent BA to whose square the rectangle is to be equal. draw ADCE, the line required. Demonstration.—By IV. 30,† EA × AD=AB2. Q. E. D. Calculation.——AC =√ AB2 + BC2; then AE AD AC— CB. AC+ CB, and PROBLEM XXXIII.-To find a square which shall be to a given square in a given ratio, as m to n. K (See IV. Prob. 11.) Analysis.--Having laid EF and FG, or m and n, contiguous on the same line EG, on EG describe a semicircle; erect the perpendicular FH to meet the semicircle in H. Join HG and HE, and produce them indefinitely. Then, drawing KI, parallel to GE, anywhere above or below GE, we always have (IV. 11, cor.) EF: FG:: HE2: HG2 : : HI2: HK2 : : HI2: HK2 in the given ratio. Construction and demonstration are manifest. Calculation.-If HK is the side of the given square, then GF: If HI is given, then EF: FG or m:n:: HI2: HK2 † I. 47. ANALYSIS BY ALGEBRA. PART I. CONSTRUCTION OF ALGEBRAIC EQUATIONS OF ONE UNKNOWN QUANTITY OF THE FIRST DEGREE, IN WHICH EACH LETTER REPRESENTS A RIGHT LINE. NOTE.-In a fraction, when the sum of the indices in the numerator exceeds the sum of the indices in the denominator by one, the fraction will represent a line; if by two, it will represent a square or surface; if by three, a cube or solid. Draw an indefinite line AF, and on it lay the lines a, b, c, and d from A to E; then x= AE. Limit. The given lines a, b, c, d, etc. may be of any number. PROBLEM II.—Given, the equation x=a+b-c+d+e—ƒ—g, to find x by construction. Draw an indefinite line AF, and on it lay all the affirmative quantities a, b, d, e successively from A to B, their extremities being marked by short lines above the line AF. Then, beginning at B, lay all the negative quantities back from B to E, their extremities being marked by short lines below the line AF. Whence we have x (c +ƒ+ g)= AB — BE a+b+d+e AE. Limit.—The lines a, b, c, d, e, etc. may be continued to any number. When the sum of the negative quantities exceeds the sum of the affirmative, E will be on the line to the left of A, and AE will be negative, or a minus quantity. PROBLEM III.-Given, the equation x= abc to find x by con ( 134 ) A struction. Here, the sum of the indices in the numerator being one more than the sum in the denominator, the result will be a line. Lay the lines represented by the letters, as in the annexed figure, AD =d, DF-a, AE b. Draw EO parallel to AB, and lay EH-e, HI= c. DE, and draw FG parallel to DE. Join HG, and abc draw IL parallel to HG. Then GL will equal X. Join de F D a For we have, by parallel lines, d:a::b:x', and e:c:: x': x or GL. By multiplying the corresponding terms, we have de: ac::b: NOTE.-The figure looks more symmetrical when EO is drawn parallel to AB, as above, but it may be drawn making any angle with EC. PROBLEM IV.-Given, the equation a2bcd2 e f g h2 i k2 l m n ' to find x by construction. Lay the lines represented by the given letters, as in the annexed figure, and draw the corresponding parallel lines, observing to take the first term, or antecedent, in the successive proportions, from the denominator, and the consequents from the numerator; and, also, that the second consequent or last term in each proportion becomes the second antecedent or third term in the next proportion. Then we have the proportions h: a:: a: x', hence x' a2 h a2b h: b :: x': x'', hence x'' etc.; ng:::; then, by multiplying the corresponding terms of these proportions, and observing that all |