ANGLE, Trisection of

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 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 C draw CE parallel to AD, meeting the extended diameter in E, and cutting the circumference of the cir-

cle 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.

Angle. 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 elementary geometry will not in general 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: it 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 the 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 DF, AD, and AF. Because AF is equal to EF, the angle EAF is equal to AEF; 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 AEF; and being likewise the double of CFD' at the circumference, the angles CFD' and AEF are equal, and, consequently, FD' and EA parallel. Now the angle CAD' being double of EAF or DAG, and the angle CAD double of DAB, the arc DCD' is double of the arcs DG and DB, which serve to complete the semicir-

cumference; wherefore this arc DCD' is two third 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 AFE' 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 CDD' is double of the angle CFD' at the circumference, and therefore double of BEF' 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 CDGD' is double of DBD', which completes the circumference. Hence the arc DDD' 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, these will be generally expressed by A, A+C, A+2C, &c.; consequently, the third part will be expressed by $\backslash frac\{1\}\{3\}A$, $\backslash frac\{1\}\{3\}A\; +\; \backslash frac\{1\}\{3\}C$, $\backslash frac\{1\}\{3\}A\; +\; \backslash frac\{2\}\{3\}C$, &c., which evidently correspond to BD, BD', and BD''. 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', and CF'' to the three points of intersection with the circumference, and the radii AD,

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

8. The simplest and most elegant solution of the trisection of an arc was indicated by Pappus, and is given in Castillon's Commentary on 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 $CBD$ 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 $CBD$ 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 $CIH$ and $KDH$ 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 $DH$. 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 $D\text{'}$, produce the perpendicular $D\text{'}K\text{'}$ to the opposite circumference $L$; and since $BD\text{'}$ is double of $D\text{'}K\text{'}$, it must be equal to the chord $D\text{'}L$, and consequently the arc $BDCD\text{'}$ is equal to $LD\text{'}MD\text{'}$. Wherefore the double of $BDCD\text{'}$, together with the interval $BL$ or $CD\text{'}$, is equal to the whole circumference; that is, the double of $DCD\text{'}$, with the double of $BD$ and $CD\text{'}$, is equal to the whole circumference; and since the double of $BD$ is $DC$, the triple of the arc $DCD\text{'}$ must complete the circumference. In the third position $D\text{'}$, produce the perpendicular $D\text{'}K\text{'}$ as before; the double of this, or the chord $MD\text{'}$, is hence equal to $D\text{'}B$, and the arc $BLD\text{'}$ equal to $DM$; consequently the double of $BLD\text{'}$, with the compound arc $BDCM$, completes the circumference; but $D\text{'}M$ being parallel to $BC$, the arc $BLD\text{'}$ is equal to $CDM$, and therefore three times the arc $BLD\text{'}$, with the arc $BDC$, or the triple of $BD$, will fill up the circumference, or the arc $DBD\text{'}$ is a third part of it.¹

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

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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 quadrissection 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 the 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 equations. 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.

The ancient geometers were only acquainted with the original division of the circumference into two, three, or five equal portions. The only subdivisions were obtained from the differences of those arcs or their continued bisection. But the very ingenious Professor Gauss has discovered a series of more complex regular polygons, which may be inscribed in a circle by elementary geometry. The expression $2^n\; +\; 1$, when a prime number, will represent the sides of the figure: for the general equation of a cosine of the section can be decomposed into quadratics of the simplest kind, which can be constructed by the repeated application of circles and straight lines. A polygon of 17 sides is the first that occurs after the pentagon; and then follow the polygons of 257, 65537, &c. (J. L.)