Cissoid of Diocles: Difference between revisions

The cissoid of Diocles is an unbounded plane curve with a single cusp, which is symmetric about the line of tangency of the cusp, and whose pair of symmetrical branches both approach the same asymptote (but in opposite directions) as a point moving along the cissoid moves farther away from the cusp. Moreover, if a circle is drawn passing through the cusp and tangent to the asymptote, then any line joining the cusp and a point on the cissoid can be extended so that it intersects the asymptote, and the length of such extension is equal to the length between the cusp and the intersection of the line with the circle.

The word "cissoid" comes from the Greek kissoeidēs : "ivy shaped" = kissos : ivy + oeidēs : shape.

Its polar equation is

$\rho =2a(\sec \theta -\cos \theta ),\qquad \qquad (1)$

where $\theta \in (-\pi /2,\pi /2)$.

Equation (1) is equivalent to the following parametric equations, viz.

$y=2a\left(\tan \theta -{1 \over 2}\sin 2\theta \right),\qquad \qquad (2)$

$x=2a\sin ^{2}\theta ,\qquad \qquad (3)$

and it is also equivalent to the following equation in rectangular coordinates,

$y^{2}={x^{3} \over 2a-x},\qquad \qquad (4)$

where $x\in [0,2a)$.

Construction

Choose a point O and a line L not passing through O. Point O and line L define the plane in which the cissoid will be drawn.

Draw a line M passing through O and perpendicular to line L. Let point P be the intersection of lines L and M. Bisect line OP at point A. Draw a circle centered at A and with radius AP. This circle is called generating circle.

Choose point Q₁ on the circle, near P. Draw line OQ₁, then extend line OQ₁ so that it intersects line L at point Q₁′. Measure the length of line Q₁Q₁′, then transfer this length over to point O: draw point O₁′ on line OQ₁′ such that the length of line OO₁′ is equal to the length of line Q₁Q₁′. Point O₁′ lies on the Diocletian cissoid.

Then choose a point Q₂ on the circle, near Q₁, but farther away from P. Draw line OQ₂, then extend line OQ₂ so that it intersects line L at point Q₂′. Measure the length of line Q₂Q₂′, then transfer this length over to point O: draw point O₂′ on line OQ₂′ such that the length of line OO₂′ is equal to the length of line Q₂Q₂′. Point O₂′ lies on the cissoid of Diocles.

Choose a point Q₃ on the circle, near Q₂ but farther away from Q₁ and from P. Repeat the steps given in the previous two paragraphs in order to obtain point O₃′ which lies on the cissoid of Diocles.

Follow up with points Q₄, Q₅, et cetera. After finding the sequence of O′ points, draw a curve through them: start at the cusp, then go through point O₁′, then through point O₂′, et cetera.

The resulting curve is one branch of the cissoid of Diocles. Reflecting this branch along the axis of symmetry OP results in its complementary branch, and both branches together form the cissoid of Diocles.

Delian Problem

The cissoid of Diocles is named after the Greek geometer Diocles who used it in 180 B.C. to solve the Delian problem: how much must the length of a cube be increased in order to double the volume of the cube?

Diocles' solution is correct, except that the solution involves an intersection of a line with a construction of a cissoid of Diocles, which cannot be accomplished by means of the simple but strict Greek rules of compass and straightedge construction.

Draw a circle centered at point A and with diameter OP. Draw a tangent line to the circle at point P and call it L. Construct a cissoid of Diocles with cusp at O, axis of symmetry OP, and asymptote L. Let B be the midpoint of the semicircular arc OP, then AB = AO = AP. Extend AB to point C such that $AC=2\cdot AB$. Draw line PC which intersects the cissoid of Diocles at point Q. Draw line OQ which intersects line AC at point M.

The cissoid of Diocles and the Delian problem.

Then $AM^{3}=2\cdot AB^{3}$, so that a cube whose side has length AM has a volume which is twice the volume of a cube whose side has length AB.

Proof: Line OQ intersects the generating circle at point K. The extension of line OQ intersects line L (the tangent) at point N. Due to the properties of the cissoid of Diocles, line OK must have the same length as line QN.

Line AC is parallel to line L = line PN. Line MN cuts the parallel lines AC and PN, therefore $\angle CMQ=\angle QNP$. Line CP cuts the parallel lines AC and PN, therefore $\angle ACQ=\angle QPN$. Also, $\angle MQC=\angle NQP$ since they are vertical angles. Then, due to angle-angle-angle congruence, $\Delta MCQ\sim \Delta NPQ$ (the two triangles are similar).

Corresponding sides of similar triangles are proportional, so

${MQ \over QN}={MC \over PN}.$

Then

${MQ+QN \over QN}={MC+PN \over PN},$

${MN \over QN}=1+{MC \over PN}.\qquad \qquad (1)$

Draw line KP. Angle OKP is a right angle, due to Thales' theorem. Angle OAM is also a right angle, since $OP\perp AC$. Then $\angle OAM=\angle OKP$. Also, $\angle KOP=\angle AOM$. Therefore $\Delta KOP\sim \Delta AOM$ due to angle-angle congruence. Then

${OA \over OM}={OK \over OP}$

$OK=OP\left({OA \over OM}\right)=2\cdot OA\cdot \left({OA \over OM}\right)\qquad \qquad (2)$

Angle OPN is a right angle, since $OP\perp PN$, so $\Delta AOM\sim \Delta PON$ (due to angle-angle congruence), and

${ON \over OM}={PN \over AM}={OP \over OA}=2$

thus

$PN=2\cdot AM$

$ON=2\cdot OM$

which means that point M is the midpoint of line ON:

$MN=ON-OM=2\cdot OM-OM=OM={ON \over 2},$

$MN=OM.\qquad \qquad (3)$

The similarity between triangles AOM and PON also implies that

${PN \over OP}={AM \over OA}$

$PN=OP\left({AM \over OA}\right)=2\cdot OA\left({AM \over OA}\right)=2\cdot AM$

i.e.

$PN=2\cdot AM.\qquad \qquad (4)$

Substitute equations (2), (3) and (4) into equation (1), which is equivalent to

${MN \over OK}=1+{MC \over PN}$

in order to obtain

${OM \over 2\cdot OA^{2}/OM}=1+{MC \over 2\cdot AM}$

$\left({OM^{2} \over 2\cdot OA^{2}}-1\right)\cdot 2\cdot AM=MC$.

From the Pythagorean theorem we deduce that

$OM^{2}=OA^{2}+AM^{2}\,$

therefore

$MC=\left({OA^{2}+AM^{2} \over 2\cdot OA^{2}}-1\right)\cdot 2\cdot MA$

$=\left({AM^{2}-OA^{2} \over 2\cdot OA^{2}}\right)\cdot 2\cdot AM$

$=\left({AM^{2} \over OA^{2}}-1\right)AM$

but $MC=AC-AM=2\cdot OA-AM$ therefore

$2\cdot OA-AM={AM^{3} \over OA^{2}}-AM,$

$2\cdot OA={AM^{3} \over OA^{2}},$

$2\cdot OA^{3}=2\cdot AB^{3}=AM^{3},$

Q.E.D.

Roulette

This curve is also a roulette. Take two congruent parabolas, set them vertex-to-vertex, and roll one along the other; the vertex of the rolling parabola will trace the cissoid.

Figure 1. A pair of parabolas — shown in black — 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 green and blue. Then the path traced by the vertex of the top parabola is it rolls is a roulette — shown in reddish color — which happens to be a cissoid of Diocles.

Define a pair of parabolas whose equations are

$y=x^{2},\$

$y=-x^{2}.\$

These parabolas face each other symmetrically across the x-axis, and are the ones shown in Figure 1. Then pick a point (a, a²) from the top parabola. The slope of the tangent to the top parabola at this point is

${d(x^{2}) \over dx}=2x.\$

The distance measured along the top parabola from point (0,0) to point (a, a²) is equivalent to the distance measured along the bottom parabola from point (0,0) to point (a, −a²). Thus, when the top parabola rolls, its point which was originally at (a, a²) will eventually come into tangential contact with the point (a, −a²) of the bottom parabola. This is true for all values of a.

When the two such points are in contact, their tangents coincide and have the same slope, which is −2 a. Let θ be the angle which the slope 2 a makes with the x-axis, so that

$\tan \theta =2a\qquad \qquad (1)$.

Then the top parabola must rotate by an angle of 2 θ in order for its point originally at (a, a²) to come into contact with the bottom parabola's point at (a, −a²).

This rotation can be performed by means of the rotation matrix R(−2 θ), viz.

$R(-2\theta )={\begin{bmatrix}\cos 2\theta &\sin 2\theta \\-\sin 2\theta &\cos 2\theta \end{bmatrix}}.$

Let the pivoting point of the rotation be point (a, a²). This pivoting vector must be subtracted from all the points on the top parabola before all these points are operated upon by the rotation matrix. Then, after the rotation is performed, this pivoting point will be at the origin, and the slope of the rotated top parabola at the origin will be −2 a, which corresponds to the slope of the point (a, −a²) of the bottom parabola. These two points must come into contact, so the rotated top parabola will be moved by vector (a, −a²) to its new position in contact with the bottom parabola. Thus the rolled top parabola will be described by

${\begin{bmatrix}\cos 2\theta &\sin 2\theta \\-\sin 2\theta &\cos 2\theta \end{bmatrix}}{\begin{bmatrix}x-a\\x^{2}-a^{2}\end{bmatrix}}+{\begin{bmatrix}a\\-a^{2}\end{bmatrix}}$

where a may be called a "rolling parameter" and x is the pre-rotated abscissa.

Applying the trigonometric identities

$\cos \,2\theta =2\cos ^{2}\theta -1\$

$\sin \,2\theta =2\sin \theta \cos \theta \$

yields a locus of points of the form

${\begin{bmatrix}2\cos ^{2}\theta -1&2\sin \theta \cos \theta \\-2\sin \theta \cos \theta &2\cos ^{2}\theta -1\end{bmatrix}}{\begin{bmatrix}x-a\\x^{2}-a^{2}\end{bmatrix}}+{\begin{bmatrix}a\\-a^{2}\end{bmatrix}},$

which, through application of equation (1), becomes

${\begin{bmatrix}2\cos ^{2}(\tan ^{-1}(2a))-1&2\sin(\tan ^{-1}(2a))\cos(\tan ^{-1}(2a))\\-2\sin(\tan ^{-1}(2a))\cos(\tan ^{-1}(2a))&2\cos ^{2}(\tan ^{-1}(2a))-1\end{bmatrix}}{\begin{bmatrix}x-a\\x^{2}-a^{2}\end{bmatrix}}+{\begin{bmatrix}a\\-a^{2}\end{bmatrix}}.$

Then, applying the identities

$\cos \tan ^{-1}x={1 \over {\sqrt {1+x^{2}}}}$

$\sin \tan ^{-1}x={x \over {\sqrt {1+x^{2}}}}$

produces the set of points of the form

${\begin{bmatrix}{2 \over 1+4a^{2}}-1&2\left({2a \over {\sqrt {1+4a^{2}}}}\right)\left({1 \over {\sqrt {1+4a^{2}}}}\right)\\-{4a \over 1+4a^{2}}&{2 \over 1+4a^{2}}-1\end{bmatrix}}{\begin{bmatrix}x-a\\x^{2}-a^{2}\end{bmatrix}}+{\begin{bmatrix}a\\-a^{2}\end{bmatrix}}$

$={\begin{bmatrix}{1-4a^{2} \over 1+4a^{2}}&{4a \over 1+4a^{2}}\\{-4a \over 1+4a^{2}}&{1-4a^{2} \over 1+4a^{2}}\end{bmatrix}}{\begin{bmatrix}x-a\\x^{2}-a^{2}\end{bmatrix}}+{\begin{bmatrix}a\\-a^{2}\end{bmatrix}}$

$={\begin{bmatrix}{(1-4a^{2})(x-a)+4a(x^{2}-a^{2}) \over 1+4a^{2}}+a\\{-4a(x-a)+(x^{2}-a^{2})(1-4a^{2}) \over 1+4a^{2}}-a^{2}\end{bmatrix}}$

$={\begin{bmatrix}{4a^{3}+4ax^{2}-4a^{2}x+x \over 1+4a^{2}}\\{2a^{2}+x^{2}-4a^{2}x^{2}-4ax \over 1+4a^{2}}\end{bmatrix}}.$

The vertex of the rolled top parabola is specified by letting x = 0, which results in the set of points which depend only on the rolling parameter a and whose coordinates are

$X(a)={4a^{3} \over 1+4a^{2}},\qquad \qquad (2)$

$Y(a)={2a^{2} \over 1+4a^{2}}\qquad \qquad (3)$.

This curve of points (X(a), Y(a)) is the roulette, and now it only remains to be shown that it is a cissoid of Diocles. The roulette is a cissoid whose equation is

$X^{2}={Y^{3} \over {1 \over 2}-Y}\qquad \qquad (4)$.

To demonstrate this, plug in the values of X and Y given in equations (2) and (3) into equation (4) and see if it leads to a tautology:

$\left[{4a^{3} \over 1+4a^{2}}\right]^{2}={\left[{2a^{2} \over 1+4a^{2}}\right]^{3} \over {1 \over 2}-\left[{2a^{2} \over 1+4a^{2}}\right]}$

${16a^{6} \over (1+4a^{2})^{2}}={\left[{8a^{6} \over (1+4a^{2})^{3}}\right] \over {1 \over 2}-{2a^{2} \over 1+4a^{2}}}$

${16a^{6} \over (1+4a^{2})^{2}}={8a^{6}/(1+4a^{2})^{3} \over {1+4a^{2} \over 2(1+4a^{2})}-{4a^{2} \over 2(1+4a^{2})}}$

${16a^{6} \over (1+4a^{2})^{2}}={8a^{6}/(1+4a^{2})^{3} \over {1 \over 2(1+4a^{2})}}={16a^{6} \over (1+4a^{2})^{2}}$

which is true for all values of a, the rolling parameter, quod erat demonstrandum.

Pedal curve

THEOREM: The pedal curve of a parabola with respect to its vertex is a cissoid of Diocles.

Proof: Any parabola can be rotated and translated so that it will end up being described by the equation

$y=ax^{2},\,$

whose slope at point (x, y) is given by the derivative

${dy \over dx}=2ax.$

Then the set of points which form the line tangent to the parabola at point (b, a b²) is

$L_{1}(b)=\{(b,ab^{2})+(u,2abu)=(b+u,ab^{2}+2abu):u\in \mathbb {R} \}$

and the set of points which form the line which is perpendicular to L₁(b) and which passes through the origin is

$L_{2}(b)=\left\{\left(v,-{v \over 2ab}\right):v\in \mathbb {R} \right\}.$

Notice that L₂(b)′s slope is −1/(2ab) which is perpendicular to L₁(b)′s slope 2ab.

The intersection L₁(b) ∩ L₂(b) can be found by setting up the following system of equations

$b+u=v,\$

$ab^{2}+2abu=-{v \over 2ab}.$

Now solve for u:

$ab^{2}+2abu=-{b+u \over 2ab}=-{1 \over 2a}-{u \over 2ab},$

$ab^{2}+{1 \over 2a}=-u\left(2ab+{1 \over 2ab}\right),$

$u={-ab^{2}-{1 \over 2a} \over 2ab+{1 \over 2ab}}={-2a^{2}b^{2}-1 \over 4a^{2}b+{1 \over b}}$

so that the abscissa v is

$v=b+u={2a^{2}b^{2} \over 4a^{2}b+{1 \over b}}$

and the ordinate is

$-{v \over 2ab}={-ab \over 4a^{2}b+{1 \over b}}.$

Thus, the point of intersection is

$\left({2a^{2}b^{2} \over 4a^{2}b+{1 \over b}},{-ab \over 4a^{2}b+{1 \over b}}\right)=\left({2a^{2}b^{3} \over 4a^{2}b^{2}+1},{-ab^{2} \over 4a^{2}b^{2}+1}\right)$

and this point belongs to the pedal curve of the parabola.

Let

$X={2a^{2}b^{3} \over 4a^{2}b^{2}+1},$

$Y={-ab^{2} \over 4a^{2}b^{2}+1}.$

Then the question is: is there a constant k which does not depend on b (but only on a) such that

$X^{2}={Y^{3} \over 2k-Y}$?

To find out, substitute the values for X and Y:

${4a^{4}b^{6} \over (4a^{2}b^{2}+1)^{2}}={-a^{3}b^{6}/(4a^{2}b^{2}+1)^{3} \over 2k+{ab^{2} \over 4a^{2}b^{2}+1}}$

Multiply the numerator and denominator on the right side by (4 a² b² + 1)³,

${4a^{4}b^{6} \over (4a^{2}b^{2}+1)^{2}}={-a^{3}b^{6} \over 2k(4a^{2}b^{2}+1)^{3}+ab^{2}(4a^{2}b^{2}+1)^{2}}$

Cross-multiply,

$8ka^{4}b^{6}(4a^{2}b^{2}+1)^{3}+4a^{5}b^{8}(4a^{2}b^{2}+1)^{2}=-a^{3}b^{6}(4a^{2}b^{2}+1)^{2}.\$

Divide by (4 a² b² + 1)²,

$8ka^{4}b^{6}(4a^{2}b^{2}+1)+4a^{5}b^{8}=-a^{3}b^{6}.\$

Distribute,

$32ka^{6}b^{8}+8ka^{4}b^{6}+4a^{5}b^{8}=-a^{3}b^{6}\$

Divide both sides by a³ b⁶,

$32ka^{3}b^{2}+8ka+4a^{2}b^{2}=-1.\,$

Subtract 4 a² b² from both sides, then factor out 8 a k on the left side,

$8ak(1+4a^{2}b^{2})=-(1+4a^{2}b^{2}).\,$

Cancel out the common factor on both sides, then solve for k,

$k=-{1 \over 8a}.$

k does not depend on b, so, for a given parabola (with fixed value of a), k is a constant, and the pedal curve is a cissoid of Diocles. Q.E.D.

Inversion

THEOREM: The inversive transform of a parabola, with the center of the inversive transformation located at the parabola's vertex, is a cissoid of Diocles.

Proof: Any parabola can be rotated and translated until it ends up being described by the equation

$y=ax^{2}.\,$

This parabola is equivalent to the set of points

$\{(x,ax^{2}):x\in \mathbb {R} \}.$

To all the points in this set apply the inversive transformation

$T_{I}:(x,y)\rightarrow \left({rx \over x^{2}+y^{2}},{ry \over x^{2}+y^{2}}\right)$

which is centered at the origin and whose radius of inversion is r. The transformed set of points is

$\left\{\left({rx \over x^{2}+a^{2}x^{4}},{rax^{2} \over x^{2}+a^{2}x^{4}}\right):x\in \mathbb {R} \right\}.$

Let

$X={rx \over x^{2}+a^{2}x^{4}},$

$Y={rax^{2} \over x^{2}+a^{2}x^{4}}.$

Now it is desired to find a value of k which is constant (not depending on x, but only on a), such that

$X^{2}={Y^{3} \over 2k-Y},$

which is the formula of a cissoid of Diocles. Plugging in the values for X and Y into the formula yields

${r^{2}x^{2} \over (x^{2}+a^{2}x^{4})^{2}}={{r^{3}a^{3}x^{6} \over (x^{2}+a^{2}x^{4})^{3}} \over 2k-{rax^{2} \over x^{2}+a^{2}x^{4}}}.$

Multiply both sides by (x² + a² x⁴)², then cancel out r² x² from both sides,

$1={ra^{3}x^{4}/(x^{2}+a^{2}x^{4}) \over 2k-{rax^{2} \over x^{2}+a^{2}x^{4}}}.$

Move the denominator from the right side of the equation to the left side,

$2k-{rax^{2} \over x^{2}+a^{2}x^{4}}={ra^{3}x^{4} \over x^{2}+a^{2}x^{4}}.$

Multiply both sides by (x² + a² x⁴),

$2k(x^{2}+a^{2}x^{4})-rax^{2}=ra^{3}x^{4}.\,$

Move the r a x² term to the right side, then factor out r a x² from both terms on the right side,

$2k(x^{2}+a^{2}x^{4})=rax^{2}(1+a^{2}x^{2}),\,$

$2kx^{2}(1+a^{2}x^{2})=rax^{2}(1+a^{2}x^{2}).\,$

Divide the common factors from both sides, then solve for k,

$k={ra \over 2}.$

For a given parabola (with fixed value of a), k is a constant. Q.E.D.

External links

• XahLee's page on the cissoid of Diocles
• MacTutor page on the cissoid of Diocles-->