Cissoid of Diocles

Ingeometry,thecissoid of Diocles(fromAncient Greekκισσοειδής(kissoeidēs)'ivy-shaped'; named forDiocles) is acubic plane curvenotable for the property that it can be used to construct twomean proportionalsto a givenratio.In particular, it can be used todouble a cube.It can be defined as thecissoidof acircleand a linetangentto it with respect to the point on the circle opposite to the point of tangency. In fact, thecurve familyof cissoids is named for this example and some authors refer to it simply asthecissoid. It has a singlecuspat the pole, and is symmetric about the diameter of the circle which is the line of tangency of the cusp. The line is anasymptote.It is a member of theconchoid of de Sluzefamily of curves and in form it resembles atractrix.

Construction and equations[edit]

Let the radius of $C$ be $a$ .By translation and rotation, we may take $O$ to be the origin and the center of the circle to be (a,0), so $A$ is $(2 a,0)$ .Then the polar equations of $L$ and $C$ are:

{\begin{aligned}&r=2a\sec \theta \\&r=2a\cos \theta.\end{aligned}}

By construction, the distance from the origin to a point on the cissoid is equal to the difference between the distances between the origin and the corresponding points on $L$ and $C$ .In other words, the polar equation of the cissoid is

r=2a\sec \theta -2a\cos \theta =2a(\sec \theta -\cos \theta ).

Applying some trigonometric identities, this is equivalent to

r=2a\sin ^{2}\!\theta \mathbin {/} \cos \theta =2a\sin \theta \tan \theta.

Let $t = tan θ$ in the above equation. Then

{\begin{aligned}&x=r\cos \theta =2a\sin ^{2}\!\theta ={\frac {2a\tan ^{2}\!\theta }{\sec ^{2}\!\theta }}={\frac {2at^{2}}{1+t^{2}}}\\&y=tx={\frac {2at^{3}}{1+t^{2}}}\end{aligned}}

are parametric equations for the cissoid.

Converting the polar form to Cartesian coordinates produces

(x^{2}+y^{2})x=2ay^{2}

Construction by double projection[edit]

Acompass-and-straightedgeconstruction of various points on the cissoid proceeds as follows. Given a line $L$ and a point $O$ not on $L$ ,construct the line $L'$ through $O$ parallel to $L$ .Choose a variable point $P$ on $L$ ,and construct $Q$ ,the orthogonal projection of $P$ on $L'$ ,then $R$ ,the orthogonal projection of $Q$ on $OP$ .Then the cissoid is the locus of points $R$ .

To see this, let $O$ be the origin and $L$ the line $x = 2 a$ as above. Let $P$ be the point $(2 a,2 at)$ ;then $Q$ is $(0, 2 at)$ and the equation of the line $OP$ is $y = tx$ .The line through $Q$ perpendicular to $OP$ is

t(y-2at)+x=0.

To find the point of intersection $R$ ,set $y = tx$ in this equation to get

{\begin{aligned}&t(tx-2at)+x=0,\ x(t^{2}+1)=2at^{2},\ x={\frac {2at^{2}}{t^{2}+1}}\\&y=tx={\frac {2at^{3}}{t^{2}+1}}\end{aligned}}

which are the parametric equations given above.

While this construction produces arbitrarily many points on the cissoid, it cannot trace any continuous segment of the curve.

Newton's construction[edit]

The following construction was given byIsaac Newton.Let $J$ be a line and $B$ a point not on $J$ .Let $∠ BST$ be a right angle which moves so that $ST$ equals the distance from $B$ to $J$ and $T$ remains on $J$ ,while the other leg $BS$ slides along $B$ .Then the midpoint $P$ of $ST$ describes the curve.

To see this,^[1]let the distance between $B$ and $J$ be $2 a$ .By translation and rotation, take $B = (-a, 0)$ and $J$ the line $x = a$ .Let $P = (x, y)$ and let $ψ$ be the angle between $SB$ and the $x$ -axis; this is equal to the angle between $ST$ and $J$ .By construction, $PT = a$ ,so the distance from $P$ to $J$ is $a sin ψ$ .In other words $a - x = a sin ψ$ .Also, $SP = a$ is the $y$ -coordinate of $(x, y)$ if it is rotated by angle $ψ$ ,so $a = (x + a) sin ψ + y cos ψ$ .After simplification, this produces parametric equations

x=a(1-\sin \psi ),\,y=a{\frac {(1-\sin \psi )^{2}}{\cos \psi }}.

Change parameters by replacing $ψ$ with its complement to get

x=a(1-\cos \psi ),\,y=a{\frac {(1-\cos \psi )^{2}}{\sin \psi }}

or, applying double angle formulas,

x=2a\sin ^{2}{\psi \over 2},\,y=a{\frac {4\sin ^{4}{\psi \over 2}}{2\sin {\psi \over 2}\cos {\psi \over 2}}}=2a{\frac {\sin ^{3}{\psi \over 2}}{\cos {\psi \over 2}}}.

But this is polar equation

r=2a{\frac {\sin ^{2}\theta }{\cos \theta }}

given above with $θ = ψ /2$ .

Note that, as with the double projection construction, this can be adapted to produce a mechanical device that generates the curve.

Delian problem[edit]

The Greek geometer Diocles used the cissoid to obtain two mean proportionals to a givenratio.This means that given lengths $a$ and $b$ ,the curve can be used to find $u$ and $v$ so that $a$ is to $u$ as $u$ is to $v$ as $v$ is to $b$ ,i.e. $a / u = u / v = v / b$ ,as discovered byHippocrates of Chios.As a special case, this can be used to solve the Delian problem: how much must the length of acubebe increased in order todoubleitsvolume?Specifically, if $a$ is the side of a cube, and $b = 2 a$ ,then the volume of a cube of side $u$ is

u^{3}=a^{3}\left({\frac {u}{a}}\right)^{3}=a^{3}\left({\frac {u}{a}}\right)\left({\frac {v}{u}}\right)\left({\frac {b}{v}}\right)=a^{3}\left({\frac {b}{a}}\right)=2a^{3}

so $u$ is the side of a cube with double the volume of the original cube. Note however that this solution does not fall within the rules ofcompass and straightedge constructionsince it relies on the existence of the cissoid.

Let $a$ and $b$ be given. It is required to find $u$ so that $u 3 = a 2 b$ ,giving $u$ and $v = u 2 / a$ as the mean proportionals. Let the cissoid

(x^{2}+y^{2})x=2ay^{2}

be constructed as above, with $O$ the origin, $A$ the point $(2 a,0)$ ,and $J$ the line $x = a$ ,also as given above. Let $C$ be the point of intersection of $J$ with $OA$ .From the given length $b$ ,mark $B$ on $J$ so that $CB = b$ .Draw $BA$ and let $P = (x, y)$ be the point where it intersects the cissoid. Draw $OP$ and let it intersect $J$ at $U$ .Then $u = CU$ is the required length.

To see this,^[2]rewrite the equation of the curve as

y^{2}={\frac {x^{3}}{2a-x}}

and let $N = (x,0)$ ,so $PN$ is the perpendicular to $OA$ through $P$ . From the equation of the curve,

{\overline {PN}}^{2}={\frac {{\overline {ON}}^{3}}{\overline {NA}}}.

From this,

{\frac {{\overline {PN}}^{3}}{{\overline {ON}}^{3}}}={\frac {\overline {PN}}{\overline {NA}}}.

By similar triangles $PN / ON = UC / OC$ and $PN / NA = BC / CA$ .So the equation becomes

{\frac {{\overline {UC}}^{3}}{{\overline {OC}}^{3}}}={\frac {\overline {BC}}{\overline {CA}}},

so

{\frac {u^{3}}{a^{3}}}={\frac {b}{a}},\,u^{3}=a^{2}b

as required.

Animation of Diocles' point-wise construction of the cissoid, using 500 randomly selected points.

Diocles did not really solve the Delian problem. The reason is that the cissoid of Diocles cannot be constructed perfectly, at least not with compass and straightedge. To construct the cissoid of Diocles, one would construct a finite number of its individual points, then connect all these points to form a curve. (An example of this construction is shown on the right.) The problem is that there is no well-defined way to connect the points. If they are connected by line segments, then the construction will be well-defined, but it will not be an exact cissoid of Diocles, but only an approximation. Likewise, if the dots are connected with circular arcs, the construction will be well-defined, but incorrect. Or one could simply draw a curve directly, trying to eyeball the shape of the curve, but the result would only be imprecise guesswork.

Once the finite set of points on the cissoid have been drawn, then line $PC$ will probably not intersect one of these points exactly, but will pass between them, intersecting the cissoid of Diocles at some point whose exact location has not been constructed, but has only been approximated. An alternative is to keep adding constructed points to the cissoid which get closer and closer to the intersection with line $PC$ ,but the number of steps may very well be infinite, and the Greeks did not recognize approximations as limits of infinite steps (so they were very puzzled byZeno's paradoxes).

One could also construct a cissoid of Diocles by means of a mechanical tool specially designed for that purpose, but this violates the rule of only using compass and straightedge. This rule was established for reasons of logical — axiomatic — consistency. Allowing construction by new tools would be like adding newaxioms,but axioms are supposed to be simple and self-evident, but such tools are not. So by the rules of classical,synthetic geometry,Diocles did not solve the Delian problem, which actually can not be solved by such means.

As a pedal curve[edit]

A pair of parabolas face each other symmetrically: one on top and one on the bottom. Then the top parabola is rolled without slipping along the bottom one, and its successive positions are shown in the animation. Then the path traced by the vertex of the top parabola as it rolls is a roulette shown in red, which is the cissoid of Diocles.

Thepedal curveof a parabola with respect to its vertex is a cissoid of Diocles.^[3]The geometrical properties of pedal curves in general produce several alternate methods of constructing the cissoid. It is the envelopes of circles whose centers lie on a parabola and which pass through the vertex of the parabola. Also, if two congruentparabolasare set vertex-to-vertex and one is rolled along the other; the vertex of the rolling parabola will trace the cissoid.

Inversion[edit]

The cissoid of Diocles can also be defined as theinverse curveof a parabola with the center of inversion at the vertex. To see this, take the parabola to bex=y²,in polar coordinate $r\cos \theta =(r\sin \theta )^{2}$ or:

r={\frac {\cos \theta }{\sin ^{2}\!\theta }}\,.

The inverse curve is thus:

r={\frac {\sin ^{2}\!\theta }{\cos \theta }}=\sin \theta \tan \theta,

which agrees with the polar equation of the cissoid above.

References[edit]

^See Basset for the derivation, many other sources give the construction.
^Proof is a slightly modified version of that given in Basset.
^J. Edwards (1892).Differential Calculus.London: MacMillan and Co. p.166,Example 3.

[1] See Basset for the derivation, many other sources give the construction.

[2] Proof is a slightly modified version of that given in Basset.

[3] J. Edwards (1892).Differential Calculus.London: MacMillan and Co. p.166,Example 3.

[1]

[2]

[3]