THE TRISECTION OF THE ARC, BY MIOHAEL HM BRENNAN, 1888, M. H. BRPJNNAN, DEVILS LAKE, DAKOTA. THE TRlISECTION OF THE ARC, BY MICHAEL H, BRENNAN, 1888, M. H. BRENNAN, DEVILS LAKE, DAKOTA. COPYRIGHT, 1888, By MICHAEL H, BRENNAN, [ALL RIGHITS RESERVED.] PREFACE. This little work is the result of a few pleasant moments, now and then, devoted to the study of the celebrated problem, "The Trisection of the Arc." I am not aware that in any of the attempts at solution, the same path has been trodden by any other person. The work is entirely original so far as my mind is con cerned. However, whether I have discovered anything not known before, I feel that these few pages will be of sufficient interest to amateurs to justify publication. It has been remarked in chapter IV that the Equilateral Hyperbola trisects the arc, and that such would appearpin the course of the discussion. By an oversight the proof of this proposition has been omitted from the imain body of the work, and is given here: Referring to figure 5 and bearing in mind that A 3=, 'a m, that the projection of T on the ellipse, is coincidently in the axes of the cone and hyperbola, that the distance from such point to the vertex is also A; it will be found that the ordinate of such point will be m. Then by the well known principle: The square of the ordinate of a hyperbola is to the rectangle of the distances from its foot to the vertices, as the square of the semi-conjugate axis is to the square of the semitransverse axis, we have: m2: 1/3m. i/-3 m::B2: m2; whence B 1/-/3m A, e =-1/. P A. Devils Lake, Dakota, June 11, 1888. M. H. B. XL \ \ ~ z, 1d~ ic CHAPTER I, that the angle O and the arc AB are actually tridivided, an c, i, the three equal angles ito whieceh O is divided. Draw the chord AB; call it m. Draw the radii OS and OT. Call the radius r and call that part of the inner radii intersected by m. y. Then the segment between m and the center will be r-y, and the segment of m, included between the inner radii, will be n-2x as will shortly appear. SAB=SOA, because SAB is measured by - of 2x and SOA, by x. NOTE.-E is the intersection of OS and AB; X, for convenience., represents the chord or the arc AS. 6 The triangle whose sides are x, x and y is similar to the triangle whose sides are r, r and x; SAB being equal to AOS and ASO common. But AOS being isosceles, SAE is isosceles; hence AE=x, and the segment of in included between the inner radii is m-2x. We now have the following proportions: X2 r: x: x y. ry=x2 or y_ r-y: m-2x:: r: x Substituting value of y in second proportion, we have: X2 r- -: n-2x::: X Xa r x-i- rm-2rx Reducing we have: 3r2g X-X3 — r_2. It will be noticed that each term represents a solid. Calling the radius 1, and substituting, we have 3x-x8 = m or x3-3x m o. Equation (1) CHAPTER II, If we apply Cardan's process to equation (1) the result will be u +v-x = -in-:im2 -+- -~n- V4m 1 -From this it will be seen that for all values of m numerically less than 2, the quantity /-_-ml will be imaginary. We know that in the equation x3'-3x + i- o, x has one real root, but it is for general values of m, an irreducible cubic, the limits of mn being, +2 and -2. The equation, serves a purpose, however, and an important one. Let us suppose that the values of x are a, b and c; then x-a- o; x-b-=o and x-c —o. Finding the product of those equations we have the following which is a more general form of the equation under discussion: x3-(a+b+ c) x2+(ab + ac+bc)x —abc= o. Placing the coefficients of this equation equal to the corresponding coefficients of equation (1) we have — a —b-c — o abi) - ac +-bc —.-3 -llabc — In, from which it will be seen that the sum of the three values of x equals o and the product equals -m, the absolute term. Thc importance of and the interest in the above will be more apparent in the next two chapters. CHAPTER III, Notwithstanding that the value of x as found by Cardlan's process, is apparently imaginary, yet there is one real value of x in equation (1), and it will be apparent before the end of this discussion that all the values of x are real. Let us suppose the real value of x to be y; then x-y o is a divisor of the equation x3 —3x-m:.^o. If we divide the equation (1) by equation x —yo, the quotient from dividing the left hand branches will be x2-+xy -+y 23. 8 Placing the quotient equal to o, we have the equation' x- -xy+y2-3-o. Equation (2) X - " y/3-y 2-. Y For y=o, x= - /3 For x-o, y= -+1i/ For y-x,wehave3x2 3; x=-+1, y +1 and-2. Equation (2) is the equation of an ellipse which is inclined from the left of the axis of ordinates at an angle of 45 and its positive parts are symmetrical with and equal to its diagonally corresponding negative parts. The semiconjugate axis is i/ ' and the semi-transverse 1/"- The curve cuts the axis of ordinates at y — + --- - and the axis of abscissas at x —=o The foci are located at 2 from the center. Figure 2 represents the ellipse; XX and YY are the axis of ordinates. Let OP=x, PQ y, PR —y. Draw QV parallel to XX and also TIU through I, the point where y intersects the semiconjugate axis. Conneet the points V and U and also R ancd S. As the curve is syimietrical anc the ordinates make an angle of 45~ with the axis, IT=QI, OP=PPI=DO=DIITZ TI —DI or PO PZ QI+PI-QP =y. Hence, PO-PZ or OZ-y. But TZZ-=X. Hence, as the point T is determined by other values of x and y, naniely, by OZ-x and TZ-y, the values of x and y are reciprocal. This nmay be algabraically shown from the form of the equation X` -3X —M=O. (More will be said in this and other chapters than is absolutely necessary for the (lemonstration but it is deemed best 9 *X- + j/_y2 _ y,Y 1 13-2 to fa thilirize t e adese ithe eeae fiur e iand t corresuponding equation.) P, Q an PR. Th valuesfor the point T are OZ,are TZ a ZS. OZ OPPK. As the values of The three roots of the equation are y, -tl.:-ayz-~-y and I-1//3 iyfy the sum' of which =o asw as shown in the equation in which a, b and c were the roots. ~ In the ellipse the three values for the point Q are OP, QP and PR. The values'for the point T are OZ, TZ and ZS. OZ'-ZN, OP~ —PK. As the. values of 10 x andly are reciprocal, PR=-ZS. That is, as x and y are reeipicail,- -PR or ZS —thc ordinate below the axis of abscissas. -y=+ idXX --- 1 Y =,/3-4x'-x +-y 1y 3 4.2- — X Y+ (+-Y) =2î/3-sx:. y+.(-y)= numerically 2lI 3 x2. 1/3-xir=y + x for the upper segment of the chord and as the lower segment is numerically equal to this, KR and NS are equal to the y's of the points Q and T respectively. That is, the portion of the chord below the transverse axis-=the portion above the axis of abscissas. The figures VQIU and TSRI are rectangles. CHAPTER IV, Figure 3 illustrates the three values of x and that they are real, also that their sum is equal to o. It will bc found that one of the values is negative and that the negative value is always equal numerically to the sum of the positive values when x is positive. The curve is produced by the intersection of the trisecting radius with the chord of the whole arc or the chord of the whole arc produced. If we suppose the point L to be fixed, the point O will describe the conchoid (that is for the given angle,) but if L is movable so -as to generate the curvc of which it is a p.)int, then:11 the conchoils intersect at O, the 1 ccnter of the circle. The distance from O to any point in the curve is 1-x2. The equation of the point according to the construction of the figure is OL=l-x-2. When x is greater than 1, the expression becornes x2-1. The equation of the point L, with LM and MO as the axes, will be found to be a biquadratic which, when reduced, gives a resulting equation identical with equation (1). If OL=l-x2, LN will equal x2 as will be readily found by inspection, the radius being taken as 1. Hence, the length of any scant passing through the centerof the circle and- terminating in the circle and the curve, will be equal to tho square of the particular x whose point of the curve is eut. Thus: PO —l-x2 for x-AP; KO1l-x2 for x=CK and OL 1-x2 for x-DL. PT —x2 for x-AP; N'K-x2 for x-CK and NL:x2 for xDL. T DE.? i j| i i __ ^ _ M 'f I ss^^. _^^y <^ -., ~F ^ Ï3~~~~ These remarks upon Figure 3 are given for what interest they may'inspire and what mental discipline there may be ini the subject, and particularly to illustrate equation (1), and show that ail of its values are real and that it and this figure are interconivertible, imathema.tically speaking. The curve bears a close resemblance to the Lemniscate of Bernomilli. The Leniniscate is the locus of the intersection of a perpe.ndicula r from the oriin on the ta. ngent to the equilaiteral hyperbolia and it will appe1ir further on that the equilateral hyperbole trisects the arc. Strictly, the point 0 is the intersection of an infinite number of conchoins —one for each possible lanl 'j]e. -- " If a perpendicular ilbe dropped from 0' the segment of OL between the center an(l the intersection of the perpendicular will be the constant portion projccting beyond the fixed line-the perpendiiculir. ^^~ M^ 13 CHAPTER V, 1-s 1ÀV in j>/ i <3 The ellipse whose equwition is x2i xy+y2 3, may be assumed to be an oblique section of a right cylinder whose dlianeter is l/ -the conjugate axis of the ellipse. The sum of the seinitransverse axis of the ellipse is 211/ 7. Hence, 1 (24 6 )- (:2) 2= the perpendicular distance from thé transverse section of the cylinder touching the upper, Vertex of the ellipse to the transverse section touching the lower vertex=1/l6 = 4. 'The Tan'le, therefore, formed by the plane of the ellipse and the plane of the base of the cylinder is the acute angle opposite 4 in the right angled triangle whose sides are 21 /6 (hypothenuse) 21/ — andl 4, respectively. Dividing by the common factor, 2, the relative length of the sides of this triangle is 1 1/ 3, i and 1/y' in which the angle referred to is opposite 1 '. 14 The triangle here discovered is the key to the solution ancd is deemed of sufficient importance to be named the MYSTIC TRIANGLE. The whole solution turns upon the relations of the sides of this triangle and the angles when named will be named by the sides tc) which they are opposite. Thus: The smaller of the two acute angles will ibe named ingle 1; the other, angle yJ 2. CHAPTER VI, Fig. 4 t PFissing a plane through the center of the ellipse perpendicular to the axis of the cylinder, a:s cirlie AB in Figure 4, the circle thus formed will have for its dliameter the conjugate axis of the ellipse and will incline to the pla-ne of the latter at an angle equal to the angle opposite 1/"' in the mystic triangle. Revolve the plane of the circle till it coincides with the plane of the ellipse. Through the center of the circle draw the lines xx and yy perpindicular to each other and at an angle of 45~ with the conjugate axis. Prolong these lines till they intersect the ellipse. Coineet the points of intersection by lines running parallel to tho transverse axis. The portion of these lines so drawn, intercepted by the circle, is the projection of the chord of the ellipse on the plane of the circle when the ellipse is in itstrue position ia the cylinder. xx and yy are the axes of ordinates by construction. yO and xO are each equal tol,/ - hence yx is 1 6; but the chord of the ellipse running parallel to the transverse axis is to its projection on thl plane of the circle as j/-: ' 1/ -. Hence 1/-: chord of circle:1/:6:1 - The chord of the circle equals 2 -~. But the sides of the angle formed by the ordinate anl abscissa being also projected-the one down the other up —on the plane of the circle and originating in the center and terminating in the ellipse, and their projection terminating in the circle; their projections are each equal to the radius (1/ ). Thus the projection of Oy is OW and that of yx is WZ, Fig. 4. The triangle found by the projection of the right-angled triangle formed by the ordinate and abscissa passing through the center of the ellipse is an equilateral triangle whose angles are, of course, each G0~. That is, the projection of the right angle whose 16 sides niake angles of 45~ with the conjugate axis of the ellipse when the plane of the ellipse makes with the plane of the circle on which it is projected an angle equal to angle opposite 1/ - in the Mystic Triangle; is 60~ or two-thirds of the right angle. The salne lLaw applies to all right angles formed by the intersection of the ordinates and abscissas at any point on the conjugate axis and the sumn of the angles projeeted on the circle is cqual to 120~ or twothirds of the sum of the supplementary angles. The author is following the path of his own investigations and desires to show not only the facts necessary to a demonstration, but also how Iand in what order the points were discovered. Applying the foregoing t(a the investigations of the relations of the ordinates and the abscissas of the ellipse to the circle revolved so that its plane coincides with the plane of the ellipse, it will be foutid that for all right angles anit integral multiples of right angles whose sides ancd the ordinates anc abscissas intersect the conjugate axes attacommnon point; the projections on the circle of the points where such ordinates and abscissas intersect the ellipse, will divide into two arcs one of which will be twice the length of the other, the arcs included between the sices of such right angles revolved so as to cut the circle. But it will be found that such is not true of angles in general. 17 CHAPTER VII, Let us suppose that CD, Figure 5,- drawn perpendicular to the conjugate axis, is the chord of onethird of a certain- arc, CI will: be one-half of the chord; its projection on the plane of the ellipse will be PI, and PI will be to CI as 1/t:1/2:: 1/3: 1. PC being perpendicular to CI, will be equal to /PI2 -CI2. But by hypothesis CI=-%x (x being the chord of one-third of the given arc); then PI= — 3 x. Substituting we have PC-l/jx..-x-i- 1/ x. That is; the ratio between x and the perpendicular connecting the extremnity of x with the ellipse is i/s. Let AT be the side of the given angle, UAT, to be trisected, TU being perpendicular to AI. Then by hypothesis TC-=x. PC being perpendicular to TC we have PT — /TC2-' pc2. l/x2 + x2 ~-2 / 6~x. That is; the distance from the extremity of the chord of the given arc to the projection on the ellipse of the point in the circle measuring one-third of the given arc from said extremity, equals -1/ X, is -~/ 6 times thu chord of one-third of the given angle. The sides of the triangle PTC are, therefore, x, l,/ 'x and -1/'6x, in which 1- '2x, or the perpendictilar, is the shortest side. Dividing out the common factor x, we have for the sides of the triangle, the relative values 1, 1~ i and ~ /-Y. Multiplying by 2 we have 2, 1/2 a-nd 1/ ~. Dividing by 1/2, we have.1/', 1 and 1/ 3. The angle at T, (PTC) being opposite the shortest side of the triangle, is less than the angle made by the plane of the ellipse and 18 the plane of the circle and is its complement. Both angles are constant for every case. Here again we have the Mystic Triangle both in PCT and PCI. TO FIND THE POINT P: Pass a plane througli the point T and perpendicultr to the plane of the circle. On this perpendicular plane lay off an angle from the intersection of the two planes, and with T as the vertex, equal to the smallest angle of the Mystic Triangle, namely angle 1. Now with T as the pivot, revolve the perpendicular plane till the upper side of the angle thus laid off intersects the ellipse on that side on which is the given arc, and from the point of intersection drop a perpendicular to the plane of the circle. Its foot will be in the circumference and(n the arc fror its foot, C, to T will be onethird of the given:arc. For, let the chord of the arc frorn T to the foot of the perpendiicular lte x; then the perpendicular PC= /-:1 x. ButP::::: PC= / i x: CI 1:: ~T:1. Whence C-I=-x- CI being one-half of the chlord CD, CD=x, anci CD:;nd TU (the chorc of the given arc) being parallel by construction, the arc DU —the arc TC-arc CD.. E. D. CHAPTER VIlI If the plane of the triangle TPC be revolved on on an axis perpendicular to the plane of the circle at T, the line TP will generate the convex surface of a tom ~,~~~~~~, i~~~~~~~~~~~~~~to 'S 61 20 cone whose vertex will be the. point T and any section of which passing through T and perpendicular to the plane of the circle will be measured by an angle equal to two right angles minus twice the smallest angle in the Mystic Triangle. When the plane of TPC in its revolution is perpendicular to the circle on the chord TU, TP"will be perpendicular to the plane of the ellipse, PTC being the complement of the angle the plane of the ellipse makes with the plane of the circle. Hence one element of the convex surface of the cone thus generated will always be perpendicular to the plane of the ellipse and will form with one-half of the chord of the given arc (~ of m),aud the distance from the middle point of m to the point at which such element of the cone cuts the plane of the ellipse, a triangle whose plane will be perpendicular to the plane of the circle and of the ellipse and whose sides will bear the ratio 1/3-, the side opposite the right angle; 1 ~, the side opposite the angle made by the intersection of the planes of the ellipse and-circle, and 1, the side opposite the angle made by the element of the cone with the plane of the circle; and each side will contain the factor one-half of m. That is, one-half of the chord of the given arc. The dist:ace from T to the intersection of TU and its projection on the ellipse is -mn and as the triangle formed lby TP" andl such intersection is the Mystic Triangle, the distance from P" to P" will be, 3 m. The distance from P" to P'' will be -l/ 3m. Hence A=:1 m. The curve formed by the intersection of the cone with the plane of the ellipse anc the element TP", in 21 the revolution of the cone, will be an Hyperbolaa plane passing through the vertex parallel to the ellipse falling within the base of the cone. The vertex of the hyperbola will be at the point where TP" is perpendicular to the plane of the ellipse and its transverse axis will be P"p P v, on which P v, the vertex of the second part of the hyperbola, will be the projection of U on the ellipse. One element of the cone which is situated at T will pass through p!". The distance between the vertex formed by the cone at T and the vertex formed by the cone produced will be four times the distance between P" and the conjugate axis. This second vertex will be in the ellipse and will also be in the axis of the cone situated at U. The vertices falling at Plv and Pv will be on the generating elements, TP and UP' respectively. It will be seen that the point required to trisect the arc is definitely fixed at the intersection of the generating element TP and the ellipse; that the point P can be found by geometrical means; that it is at the intersection of three surfaces, namely, the convex surface of a cone, the convex surface of the right cylinder of which the circle TU is a transverse section, and the plane of the ellipse formed by passing a plane through the cylinder which will contain the diameter IO and at an angle with the circle equal to the greater acute angle in the Mystic Triangle. If the operation of passing the planes and surfaces as above stated is actually performed the intersections will be: An ellipse, and an hyperbola. The hyperbola will cut the ellipse at P and p, and two other points-one at Plv and one a little to the left of Piv as shown in the figure. CHAPTER IX X3 - PX+Q=0. Equation (3) In every cubic equation of the form X3 -PX4- Q =0, when Q is no greater than it i/X, the Yvalue of X is equal to the chord of one-third of an arc subtended by Q in a circle whose radius is ~1/ 6p. Suppressing the equation as in chapter III, we have X2+ XY + Y- - P. Equation (4) X - /p py2-iY Equation (4) is the equation of an ellipse similar to the one described in Chapter III, and all that was said of equation (2) will apply to equation (4) Hence when +X +Y, X2 -- Y2 the square of the semiconjugate axis = the square of the radius of the circle cut from the cylinder. WhenX-Y, 3X2 -P; X - + jiJP X2+ Y - tP= R2 B2. R= - / 6P_ =B. As the greatest value Q can have in the circle is twice the radius, and the radius is ~ 1/6P, Q maximum t/,6P-. Q. E. D. When Q is less than t1l/, the value of X as found by Cardan's process will be imaginary. It has long been known that in Cardan's irreducible case all the 23 values of X are real. The celebrated Bombelli declared that the irreducible case in Cardan could be solved only by the "trisection of an arch. " By what process he arrived at such a conclusion the writer of this does not know; but it is evident from the foregoing that the irreducible case is, at least embraced in the problem of the trisection of the arc, and that the two problems are identical. ERRATA. In Preface for "-i /- m." read "~ 1/S mr." On last line of page 7 "Y" should be "Y2., On page 8, line 10, for "'45" read "45o." On page 8 line 15, for "/ ' read "-'T)" On page 8, line 18, for "axis" read "axes." On page 18, in first of last four lines, insert "00' after Iperpendicular."' Omit last paragraph of page 16. It involves an error and is not necessary to the completeness of the work. On 9th line from bottom of page 22, for "~l/,p" read '.-.,