PROBLEM XI. To find a square which shall be to a given square as a given line to a given line. Let AC be the given D square, and M and N the given lines. Upon the indefinite line EG, take EF=M, and FG=N; upon EG as a diameter describe A CM N B a semicircle, and at the point F erect the perpendicular FH. From the point H, draw the chords HG, HE, which produce indefinitely upon the first, take HK equal to the side AB of the given square, and through the point K draw KI parallel to EG; HI will be the side of the square required. For, by reason of the parallels KI, GE, we have HI : HK :: HE: HG; hence, HI2: HK2 :: HE2 : HG2: but in the right angled triangle EHG, the square of HE is to the square of HG as the segment EF is to the segment FG (Prop. XI. Cor. 3.), or as M is to N; hence HI: HK2:: M: N. But HK=AB; therefore the square described upon HI is to the square described upon AB as M is to N. PROBLEM XII. Upon a given line, to describe a polygon similar to a given polygon. Let FG be the given line, and AEDCB the given polygon. In the given polygon, draw the diagonals AC, AD; at the point F make the angle GFH= BAC, and at the point G the angle FGH=ABC; the lines FH, GH will cut each other in H, and FGH will be a triangle similar to ABC. In the same manner upon FH, homologous to AC, describe the triangle FIH similar to ADC; and upon FI, homologous to AD, describe the triangle FIK similar to ADE. The polygon FGHIK will be similar to ABCDE, as required. For, these two polygons are composed of the same number of triangles, which are similar and similarly situated (Prop. XXVI. Sch.). PROBLEM XIII. Two similar figures being given, to describe a similar figure which shall be equivalent to their sum or their difference. Let A and B be two homologous sides of the given figures. Find a square equivalent to the sum or to the difference of the squares described upon A and B ; let X be the side of that square ; then will X in the figure required, be the side which is homologous to the sides A and B in the given figures. The figure itself may then be constructed on X, by the last problem. A B X For, the similar figures are as the squares of their homologous sides; now the square of the side X is equivalent to the sum, or to the difference of the squares described upon the homologous sides A and B; therefore the figure described upon the side X is equivalent to the sum, or to the difference of the similar figures described upon the sides A and B. PROBLEM XIV. To describe a figure similar to a given figure, and bearing to it the given ratio of M to N. Let A be a side of the given figure, X the homologous side of the figure required. The square of X must be to the square of A, as M is to N: hence X will be found by (Prob. XI.), and knowing X, the rest will be accomplished by (Prob. XII.). A X PROBLEM XV. To construct a figure similar to the figure P, and equivalent to the figure Q. Find M, the side of a square equivalent to the figure P, and N, the side of a square equivalent to the figure Q. Let X be a fourth proportional to the three given lines, M, N, AB; upon the side X, homologous to AB, P B describe a figure similar to the figure P; it will also be equivalent to the figure Q. For, calling Y the figure described upon the side X, we have P : Y :: AB2 : X2; but by construction, AB : X :: M : N, or AB2: X2:: M2: N2; hence P: Y:: M2: N2. But by construction also, M2=P and N2=Q; therefore P: Y :: P: Q; consequently Y=Q; hence the figure Y is similar to the figure P, and equivalent to the figure Q PROBLEM XVI. To construct a rectangle equivalent to a given square, and having the sum of its adjacent sides equal to a given line. Let C be the square, and AB equal to the sum of the sides of the required rectangle. Upon AB as a diameter, describe a semicircle; draw the line DE parallel to the diameter, at a distance AD from it, equal to the side of the given square C; from the point E, where the parallel cuts the circumference, draw EF perpendicular to the diameter; AF and FB will be the sides of the rectangle required. For their sum is equal to AB; and their rectangle AF.FB is equivalent to the square of EF, or to the square of AD; hence that rectangle is equivalent to the given square C. Scholium. To render the problem possible, the distance AD must not exceed the radius; that is, the side of the square C must not exceed the half of the line AB. PROBLEM XVII. To construct a rectangle that shall be equivalent to a given square, and the difference of whose adjacent sides shall be equal to a given line. Suppose C equal to the given square, and AB the difference of the sides. Upon the given line AB as a diameter, describe a semicircle: at the extremity of the diameter draw the tangent AD, equal to the side of the square C; through the point D and the centre O draw the secant DF; then will DE and DF be the adjacent sides of the A rectangle required. For, first, the difference of these sides is equal to the diameter EF or AB; secondly, the rectangle DE, DF, is D equal to AD2 (Prop. XXX.); hence that rectangle is equivalent to the given square C. PROBLEM XVIII. To find the common measure, if there is one, between the diagonal and the side of a square. Let ABCG be any square whatever, and AC its diagonal. G D E We must first apply CB upon CA, as often as it may be contained there. For this purpose, let the semicircle DBE be described, from the centre C, with the radius CB. It is evident that CB is contained once in AC, with the remainder AD; the result of the first operation is therefore the quotient 1, with the remainder AD, which latter must now be compared with BC, or its equal AB. A F B We might here take AF-AD, and actually apply it upon AB; we should find it to be contained twice with a remainder: but as that remainder, and those which succeed it, con tinue diminishing, and would soon B only upon lines which remain always of the same magnitude. The angle ABC being a right angle, AB is a tangent, and AE a secant drawn from the same point; so that AD : AB :: AB AE (Prop. XXX.). Hence in the second operation, when AD is compared with AB, the ratio of AB to AE may be taken instead of that of AD to AB; now AB, or its equal CD, is contained twice in AE, with the remainder AD; the result of the second operation is therefore the quotient 2 with the remainder AD, which must be compared with AB. Thus the third operation again consists in comparing AD with AB, and may be reduced in the same manner to the comparison of AB or its equal CD with AE; from which there will again be obtained 2 for the quotient, and AD for the remainder. Hence, it is evident that the process will never terminate; and therefore there is no common measure between the diagonal and the side of a square: a truth which was already known by arithmetic, since these two lines are to each other :: √2:1 (Prop. XI. Cor. 4.), but which acquires a greater degree of clearness by the geometrical investigation. |