ANGLE

ANGLE (TRISECTION OF). The attempts of the Greek mathematicians to Double a Cube and to Trisect an Angle, were their first steps beyond the limits

Trisection of Elementary Geometry. They soon perceived that such problems cannot be solved by any combination of mere straight lines or circles. To this conclusion they were led directly by the application of Geometrical Analysis, a beautiful instrument of discovery which Plato had recently invented or improved. Their investigations pointed at some curves of a higher order than the circle, and opened to them a wide and interesting field of research.

The analysis of the trisection of an angle, conducted in two different ways, terminates in the construction of the Conchoid, a complex curve which was first proposed by Nicomedes. As the subject is very curious, and throws great light on the theory of angular magnitude, we shall here not only give both the ancient methods of investigation, but subjoin a third which is due to the sagacity of Newton.

1. Let it be required to trisect the angle BAC or the arc BC. Suppose the thing already done, and the angle BAD to be the third part of the given angle. From the point CE draw CE parallel to AD, meeting the extended diameter in E, and cutting the cir-

cumference of the circle in the point F; join FD and FA. It is obvious, that the angle BAD is the half of DAC, the remaining part of the whole angle, and therefore, equal to the angle DFC at the circumference. But AD and EC being parallel, the angle BAD is equal to AEF, which is hence equal to DFC; and consequently FD is parallel to EB. Wherefore, the arc BD is equal to GF, and the angle BAD equal to GAF. The angle AEF is thus equal to EAF, and hence the side EF is equal to AF, the radius of the circle.

To solve the problem, therefore, it would be requisite to inflect from C a straight line CFE, such that the portion FE, intercepted between the circumference and the diameter or its extension, should be equal to the radius of the circle. The radius AD, drawn parallel to this inflected line, CE would cut off an angle BAD, which is the third part of the given angle BAC.

But it is clear, that Elementary Geometry will not furnish the means of inflecting CE, according to the required conditions. This must be done either tentatively, that is, by repeated trials, or by the application of a curve, so constituted, that every straight line drawn from the pole C, to the directrix BG, shall have the portion EF, intercepted by the curve equal to AB. This curve is, from its general shape or resemblance to a conch or shell, named the Conchoid; consists of two branches, one above the directrix called the interior conchoid, and the other

below it, called the exterior conchoid. The conchoid being described, will, by its intersection with the circumference of the circle, give the point F, and consequently the position of trisecting line AD'. But such a complex curve must cut the circumference in more points than one, and consequently the problem of angular trisection, viewed in its generality, admits of several answers. In fact, there are always three distinct positions of the inflected line CE, which will fulfil the conditions of the problem.

It is curious to examine these different positions of the inflected line. Draw AD' parallel to the second position CE', and join D'F, AD', and AF'. Because AF' is equal to E'F', the angle E'AF' is equal to AE'F'; and, consequently, the exterior angle AFC is the double of either of these. But CAF' being an isosceles triangle, AFC is equal to ACF; which again is equal to the alternate angle CAD'; wherefore, CAD' is the double of the angle AE'F'; and being likewise the double of CFD' at the circumference, the angles CFD' and AE'F' are equal; and, consequently, F'D' and E'A parallel. Now the angle CAD' being double of E'AF' or D'AG, and the angle CAD double of DAB, the arc DCD' is double of the arcs D'G and DB, which serve to com-

plete the semicircumference; wherefore, this arc DCD' is two-thirds parts of the semicircumference, or one-third of the whole circumference.

In the third position AD'' of the trisecting line, draw CF'' parallel to it, and join AF'' and D''F''. The isosceles triangles D''AF'' and AF''E'' have equal vertical angles; and, consequently, the angles at their base are likewise equal; wherefore, AF''D'' is equal to the alternate angle F''AE'', and the chord D''F'' parallel to the diameter BG. But the reverse angle CAD'', standing on the arc CD''D'', is double of the angle CF''D'' at the circumference, and, therefore, double of BE''F'' or of BAD''; and the angle CAD being, by the first construction, likewise double of BAD, the reverse angle CAD'', together with CAD, must be double of BAD'' and BAD, or the arc CD''GD'' is double of DBD'', which completes the circumference. Hence the arc DD''D'' is two-thirds of the circumference.

It thus appears that the construction of the problem assumes three different aspects, and that the trisecting lines, to which a close analogy conducts us, mutually divide the whole circuit into equal portions. These results are perfectly conformable with the theory of angular magnitude. For if A denote the arc BC, and C the whole circumference, this arc

Trisection of. will be generally expressed by $A,\; A+C,\; A+2C,\; \backslash dots$; consequently, the third part will be expressed by $\backslash frac\{1\}\{3\}A,\; \backslash frac\{1\}\{3\}A+C,\; \backslash frac\{1\}\{3\}A+2C,\; \backslash dots$, which evidently correspond to $BD,\; BD\text{'},\; BD\text{'}\text{'},\; \backslash dots$. But any farther extension of this progression only brings the trisecting line back into its former positions.

To solve completely, therefore, the problem of the trisection of an angle, from the pole $C$, on either side of the directrix $BG$, with a measure equal to the radius of the circle, describe the exterior and the interior or nodated conchoid; draw $CF,\; CF\text{'},\; CF\text{'}\text{'}$ to the three points of intersection with the circumference,

and the radii $AD,\; AD\text{'},\; AD\text{'}\text{'}$, parallel to these will mark the triple section of the angle $BAC$.

It may be perceived that the exterior branch of the conchoid cuts the under semicircle in another point besides $F$. This occurs in the extension of the radius $CA$, or where the diameter, passing through $C$, the extremity of the original arc, meets the opposite circumference; the portion of the inflected line, intercepted below $BG$ by the conchoid, being evidently equal to the radius. The fourth intersection, however, affords no real solution, but only exhibits the amount of repeated division, as completing the arc itself.

2. But another analysis leads to a similar result. Let the angle $BAD$ as before be third part of $BAC$; draw $BC$ perpendicular to $AB$ and $CE$ parallel to it, meeting $AD$ produced in $E$. The right angle $DCE$ would be contained in a semicircle having $DE$ for its diameter; join $C$ with the centre $H$, and the triangle $CHE$ being isosceles, the exterior angle $CHA$ is double of $CEH$, or of the alternate angle $BAD$; and, therefore, equal to the remaining portion $CAD$ of the divided angle $BAC$. Whence the triangle $ACH$ is isosceles, and the side $CH$ equal to $HC$, or the diameter $DE$ must be double of $AC$.

The construction of the problem is thus reduced to the drawing from the vertex to the given angle, a straight line, $ADE$, such that the part $DE$ intercepted between the perpendiculars $BC$ and $CE$, shall be equal to the double of $AC$. This can only be done by describing a conchoid from the pole $A$ to the directrix $BC$, and with the double of $AC$ as the measure; the intersection of the curve with the perpendicular $CE$, will determine the position of the trisecting line $ADE$. The exterior branch of the conchoid, will cut the perpendicular in the point $E$, and the

interior or nodated branch will meet and cross it at the two points $E\text{'}$ and $E\text{'}\text{'}$. The radiating lines $AE,\; AE\text{'},\; AE\text{'}\text{'}$, or its extension $AE\text{'}\text{'}$, will indicate the complete trisection of the angle $BAC$. These lines will be found, as in the first construction, to make angles with each other that are equal to the thirds of an entire circuit. It may be worth while to examine the several cases.

In the second position $D\text{'}AE\text{'}$ of the trisecting line, draw $CH\text{'}$ to the middle point. Because the triangle $EHC$ is isosceles, its exterior angle $CH\text{'}A$ is double of $CEH\text{'}$, or of the angle $BAD\text{'}$; but $H\text{'}CA$ being also an isosceles triangle, $CH\text{'}A$ is equal to $CAH$, and consequently double of $BAD$. Add $CAD$, which is double of $BAD$, and the compound angle $DAE\text{'}$ is double of $DAD\text{'}$, which would complete two right angles; whence $DAE\text{'}$ is two-thirds of two right angles, or one-third of a whole circuit.

In the third position, $AE\text{'}\text{'}D\text{'}\text{'}$ of the trisecting line, or rather its extension $AE\text{'}\text{'}$, draw $CH\text{'}\text{'}$ to bisect $E\text{'}\text{'}D\text{'}\text{'}$. The triangles $CH\text{'}\text{'}D$ and $ACH\text{'}\text{'}$ are then isosceles, and consequently the angle $CAD\text{'}\text{'}$ or $CH\text{'}\text{'}A$ is double of $CD\text{'}\text{'}A$; but $CAE$ is likewise double of $CEA$, and therefore the combined angle $D\text{'}\text{'}HE$ is double of the angles $CD\text{'}\text{'}A$ and $CEA$. Now this angle $D\text{'}\text{'}HE$, together with the two angles $CD\text{'}\text{'}A$ and $CEH$, is evidently equal to the exterior angle $DC\text{'}E$ or a right angle. Whence $D\text{'}\text{'}AE$ is two-thirds of a right angle, or one-third of two right angles, and therefore the adjacent angle $DAE\text{'}\text{'}$ is two-thirds of two right angles, or one-third of a whole circuit.

3. The simplest and most elegant solution of the trisection of an arc is due to modern science, and first appeared in Newton's Universal Arithmetic.

The problem is there reduced to the combination of the circle with a certain kind of hyperbola. But the general property of the directrix, which belongs to all the conic sections, or the lines of the second order, affords the readiest mode of investigation. Let the arc BD be the third part of BDC. Complete the circle, and draw the chords BD, BC, and CD. The arc CD is evidently double of BD, and therefore, the angle CBD is double of BCD. Bisect the angle CKD, by the straight line BH, let fall the extended perpendicular, and draw the parallel DK. The triangle CHB is evidently isosceles, and HI bisects the base CB. But the triangle HBD having its vertical angle at B bisected, the side CB is to BD, as the segment CH of the base to HD, that is, since the triangles CHI and DHK are similar as CI to KD; wherefore CB being the double of CI, BD is likewise double of DK. The ratio of the distances BD and DK is thus given, while the point B is given, and the straight line IH given in position. Whence, from the theory of Lines of the Second Order, the locus of the point of section D is an hyperbola, of which B is a focus, and IH a directrix, with the determining ratio of two to one. Let this construction be made, and the arc CDB, is trisected in D. For since BD is, from the property of the curve, double of DK, it is evident that BC is to CI as BD to DK; and the triangles CHI and KBH being similar, and CI to CH, as DK to DH, it follows that BC is to CH as BD to DH, or alternately BC is to BD as CH to DK. Wherefore the vertical angle CBD is bisected by BH, or the angle CBD is double of CBH or of BCD; and consequently the arc CD is double of BD, or BD itself is the third part of the whole arc CDB.

But the opposite branch of the hyperbola, which passes through C, also comes into play; and the intersection of these two branches with the circle, assigns three different positions of the point D, separated from each other by intervals equal to the third of the whole circumference. Thus, in the second position D', produce the perpendicular D'K' to the opposite circumference L; and since BD' is double of D'K', it must be equal to the chord D'L, and consequently the arc BDCD' is equal to D'MD'. Wherefore the double of BDCD', together with the interval BL or CD, is equal to the whole circumference; that is, the double of DCD' with the double of BD and CB' is equal to the whole circumference; and since the double of BD is DC, the triple of the arc DCD' must complete the circumference. In the third position D'', produce the

perpendicular DK as before; the double of this or the chord BD is hence equal to D'M, and the arc BLD equal to DM. Consequently the double of BLD, with the compound arc, BDCM completes the circumference; but D'M, being parallel to BC, the arc BLD is equal to CDM, and therefore three times the arc BLD with the arc BDC, or the triple of BD will fill up the circumference, or the arc DBD is a third part of it.

The trisection of innumerable arcs described on the same chord is rendered very conspicuous, by combining the separate branches of two hyperbolas that have the determining ratio of two to one, and their foci situate in the extremities of the given line. Thus, let the chord BC be trisected at the points N and O, and from B and C, as distinct foci, and in the determining ratio of two to one, describe branches of independent hyperbolas. All the arcs erected on BC, are each of them divided by those curves into three equal portions. These arcs, as they flatten, approach to the trisected chord CNOB, on the one hand; and as they become enlarged, they constantly tend, on the other, to the complete circumference which the asymptotes of the hyperbola, making angles on each side of the axis equal to two-thirds of a right angle, would themselves trisect.

It may be observed in general, that the section of an arc or angle admits of as many different answers as the number of divisions proposed. Thus, the quadrisection would give four distinct results, and a quinsection would

involve no fewer than five separate products. Nay, the bisection itself of an arc, though within the limits of the most elementary geometry, yet brings out a double result. Thus, the arc CDB is bisected by perpendicular or diameter DID', which not only gives BD

for the half of that arc, but also BDCD', or the same half arc augmented by a semicircumference.

These conclusions agree with the results derived from the general theory of functions. The expression for the sine of a multiple arc, is always an equation of corresponding dimensions; which therefore admits of as many distinct roots as the index contains units. (D.)