Dandelin Spheres Theory

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Dandelin Spheres
This image shows a head floating on the ocean surface with a funny cone-shaped hat. A round fish is kissing the ocean surface under the water in the hat. The hat intersects the ocean surface in a "Conic Section", which in the image is an ellipse. This image is an example of Dandelin Spheres structure.

Basic Description

Mathematicians see this image more from a mathematical perspective. We can view the water level inside the cone as being an ellipse created by the intersection of a cone and a plane, and the fish and the man's head as being spheres that are tangent to the cone and to the ellipse. The two spheres appear to touch the two foci of the ellipse.

In geometry, the Dandelin spheres are one or two spheres that are tangent both to a plane and to a cone that intersects the plane. The intersection of the cone and the plane is a conic section, and the point at which either sphere touches the plane is a focus of the conic section, so the Dandelin spheres are also sometimes called focal spheres. See Figure 1.

Since the Dandelin Spheres are created by conic sections, the types of Dandelin Spheres can be categorized by different types of conic sections:

Here is a video helps to understand the Dandelin Spheres.

This video shows the effect of light, which causes the ball's shadow and creates different Dandelin Spheres.

At 00:04, it creates dandelin spheres tangent to a circle.

From 00:05 to 00:09, with the movement of the light, it shows dandelin spheres tangent to an ellipse.

From 00:10 to 00:14, it shows dandelin spheres tangent to a pair of hyperbola.

At 00:16, it shows dandelin spheres tangent to a parabola.

History

Dandelin Spheres are named after the Belgian mathematician and military engineer Germinal Pierre Dandelin (1794–1847). Adolphe Quetelet(1796–1874), famous Belgian astronomer, mathematician, statistician and sociologist, is sometimes given partial credit as well. Conic sections are defined by the intersection of a plane with a cone. Dandelin published an article, in which he showed the importance of the intersections created by spheres tangent to the cone and a conic section. He also gave an elegant proof that the spheres are tangent to the conic section at its foci.

The Dandelin spheres can be used to prove at least two important theorems. Both of those theorems were known for centuries before Dandelin, but he made it easier to prove them.

A More Mathematical Explanation

Note: understanding of this explanation requires: *Geometry, Algebra and Basic Knowledge of Conics Sections

Two Theorems

Sum of Distances to Foci Property: the first theorem proven by Dandelin Sphe [...]

Two Theorems

Sum of Distances to Foci Property: the first theorem proven by Dandelin Spheres is that a closed conic section (i.e. an ellipse) is the locus of points such that the sum of the distances to the foci is constant. This was known to Ancient Greek mathematicians such as Apollonius of Perga, but the Dandelin spheres facilitate the proof. See Figure 1
Focus-Directrix Property: the second theorem proven by Dandelin Spheres is that for any conic section, the distance from the focus is proportional to the distance from the directrix, the constant of proportionality being called the eccentricity. Again, this theorem was known to the Ancient Greeks, such as Pappus of Alexandria, but the Dandelin spheres facilitate the proof.
The directrix of a conic section can be found using Dandelin's construction. Each Dandelin sphere intersects the cone at a circle; let both of these circles define their own planes. The intersections of these two planes with the conic section's plane will be two parallel lines; these lines are the directrices of the conic section. However, a parabola has only one Dandelin sphere, and thus has only one directrix.
Using the Dandelin Spheres, it can be proved that any conic section is the locus of points for which the distance from a point (focus) is proportional to the distance from the directrix. Neither Dandelin nor Quetelet used the Dandelin spheres to prove the focus-directrix property. The first to do so was apparently Pierce Morton in 1829. The focus-directrix property is essential to proving that astronomical objects move along conic sections around the Sun.
Germinal Pierre Dandelin employs spheres inscribed in a cone which touch the intersecting plane in two points, the foci of the conic section. In what follows, both of Dandelin's proofs are presented.
This will be shown in Explore Different Dandelin Spheres.

Sum of Distances to Foci Property

Explore Sum of Distances to Foci Property for different dandelin spheres.

Before we start to explore Sum of Distances to Foci Property for different dandelin spheres, we need to know that if two line sections which start from the same point and both tangent to a sphere, they are equal.

The image Proof of the Sphere shows a sphere centered at point $O$.
Proof of the Sphere
Line $PA$ is tangent to the sphere at point $A$, and line $PB$ is tangent to the sphere at point $B$. Because the lines are tangent to the sphere, $\angle OAP$ and $\angle OBP$ are right angles. The two triangles $\triangle OAP$ and $\triangle OBP$ share the side $OP$. Sides $OA$ and $OB$ are radii of the sphere, so $OA=OB$. Since they both own a right angle, one equal length leg and same hypotenuse, the two triangles are congruent triangles.

Thus, $AP=BP$.

Imagine all such lines that start at P and touch the sphere. All these lines will form a conical cap of the sphere. It is then easy to see that the distances from $P$ to all the points of contact at the sphere are equal. The base of a cone is a circle, with all the points on that certain equidistant from the top vertex. This means $AP=BP$.

Use this Applet to play with Dandelin Spheres.

Circle

Sum of Distances to Foci Property for dandelin spheres tangent to a circle.

Figure 2

In this part we will show how Dandelin proved the Sum of Distances to Foci Property:

In Figure 2, when the plane intersects all generators of the cone, it is possible to inscribe two spheres which will touch the conical surface and the plane.

The intersection plane is a circle centered at $F$, showing in red line. The upper sphere centered at point $C$ touches the cone surface in a circle $k$ and the circle at its center $F$. The lower sphere centered at point $C'$ touches the cone surface in a circle $k'$ and also touches the circle at the center $F$.

The arbitrary chosen generating line $TT'$ intersects the circle $k$ at a point $T$, the circle $k'$ at a point $T'$, and the intersection curve at a point $P$.

We see that points $T$ and $F$ are tangency points of the upper sphere, and points $T'$ and $F$ are the tangency points of the lower sphere.

Proof:

Because $TP$ and $PF$ both tangent to the upper sphere, thus

Eq. 1         $TP = PF$.

By using the same method, we can get:

Eq. 2         $T' P = PF$.

From Eq. 1 and Eq. 2, we can get:

Eq. 3         $TP + T' P = TT' = 2 PF$.

Since point $P$ is an arbitrary point on the conic section, and the length of $TT'$ is constant, $PF$ must be constant. This proves Sum of Distances to Foci Property in circle.

Ellipse

Sum of Distances to Foci Property for dandelin spheres tangent to an ellipse.

Figure 3. Created by Wikipedia

In Figure 3, when a plane intersects all generators of the cone, it is possible to inscribe two spheres which will touch the conical surface and the plane. In this section I am going to introduce how to prove Sum of distances to foci property for dandelin spheres in this shape.

The upper sphere $G_1$ touches the cone surface in a circle $k_1$ and the plane at a fixed point $F_1$. The lower sphere$G_2$ touches the cone surface in a circle $k_2$ and the plane at another fixed point $F_2$.

The arbitrarily chosen generating line $S P_2$ intersects the circle $k_1$ at a point $P_1$, the circle $k_2$ at a point $P_2$ and the intersection curve $E$ at a point $P$.

We see that points $P_1$ and $F_1$ are tangency points of the upper sphere and points $P_2$ and $F_2$ are tangency points of the lower sphere.

Proof:

Since sphere $G_1$ is tangent to the cone and the intersection $k_1$ of the sphere and the cone is a circle containing point $P_1$, line $P_1 P$ is tangent to sphere $G_1$. The sphere is also tangent to the conic section at point $F_1$, so the line $F_1 P$ on the conic section is also tangent to the sphere. Because of the proof we gave above in Proof of the Sphere, two tangent line sections of the same sphere are equal.

Eq. 1         $P_1 P = F_1 P$.

Using the same method, we can get:

Eq. 2         $P_2 P = F_2 P$.

Since points $P$, $P_1$, and $P_2$ are on one line,

Using Eq. 1 $+$ Eq. 2, we can get:
Eq. 3         $P_1 P + P_2 P = F_1 P + F_2 P$.
Eq. 4         $P_1 P_2 = F_1 P + F_2 P$.

$P_1 P_2$ is the line section on the cone between circle $k_1$ and circle $k_2$.

1. Spokes from the upper sphere’s hat band are all equal because they are tangent lines sharing a far point.
2. Spokes from the lower sphere’s belt are all equal length for the same reason.
3. Take away hat spokes from belt spokes and you're left with lamp shade spokes which are all equal.

$P_1 P_2$ is constant, and $F_1 P + F_2 P$ is constant too. This proves Sum of Distances to Foci Property in ellipse.

Hyperbola

Sum of Distances to Foci Property for dandelin spheres tangent to a hyperbola.

Figure 4

When the intersecting plane is inclined to the vertical axis at a smaller angle than that of the generator of the cone, the plane $\pi$ cuts both cones creating a hyperbola, which is shown in Figure 4.

Inscribed spheres centered at $C'$ and $C$ touch the plane from the same side at points $F'$ and $F$, and the cone surface at circles $k'$ and $k$.

The generator intersects the circles $k'$ and $k$ at points $T'$ and $T$, and the intersection curve at the point $P$.

By rotating the generator around the vertex $V$ by 360 degrees, the point $P$ will move around and trace both branches of the hyperbola.

Proof:

Point $P$ is an arbitrary point on the conic section. Line $T' P$ goes through the two spheres, and intersects the upper sphere at point $T'$. The upper sphere intersects the plane containing the conic section at point $F'$, and intersects the cone in a circle $k'$. The lower sphere intersects the plane containing the conic section at point $F$, and intersects the cone in a circle $k$. Using the same technique shown in Proof of the Sphere, since Points $T'$ and $F'$ are on the sphere centered at $C'$, we can get:

Eq. 1         $F' P=T' P$.

Since Points $T$ and $F$ are on the sphere centered at $C$,

Eq. 2         $F P=T P$.

Since the planes of circles $k'$ and $k$ are parallel, all generating segments from $k'$ to $k$ are of equal length.

Since points $P$, $T'$, and $T$ are on one line,

Using Eq. 2 $-$ Eq. 1, we can get:
Eq. 3         $|F P - F' P|=|T P - T' P|$,
Eq. 4         $T' T = |F P - F' P|$.

$T' T$ is the line section on the cone between circle $k'$ and circle $k$.

$T' T$ is constant, so $|F P - F' P|$ is constant too. This proves Sum of Distances to Foci Property in hyperbola.

Parabola

Sum of Distances to Foci Property for dandelin spheres tangent to a parabola.

Figure 5

See Figure 5. An inscribed sphere, centered at $C$, touches the plane at point $F$ and the cone surface at circle $k$.

The generator intersects the circle $k$ at point $T$, and the intersection curve at point $P$.

By rotating the generator around the vertex $V$ by 360 degrees, the point $P$ will move around the conic section.

The sphere is centered at point $C$ situated below the cone vertex $V$. The sphere touches the surface of the cone from the inside in a circle $k$ and it also touches the intersecting plane at a point $F$. In addition to the intersecting plane there is also another plane displayed on the image. It is the plane in which circle $k$ is embedded. The intersection of this plane with the intersection plane is a line $p$.

$P$ is an arbitrary point on the conic section.

Proof:

Point $T$ is the intersection point of the line $PV$ with circle $k$. The point $Q$ is the perpendicular projection of point $P$ to the plane of circle $k$. The point $S$ is the foot of a perpendicular from point $P$ to line $p$.

This creates two triangles: $\triangle PQT$ and $\triangle PQS$.

As we showed in Proof of the circle, because points $F$ and point $T$ are on the sphere, we know that:

Eq. 1         $|PF| = |PT|$.

In this case we cannot use Eq. 1 directly to get the answer. If we want to prove Sum of Distances to Foci Property, we need to prove that $|PT| = |PS|$.

Lines $PT$ and $PS$ are hypotenuses of two right triangles $\triangle PQT$ and $\triangle PQS$. These two triangles are congruent since they have one identical side $PQ$ and two congruent angles.

Their right angles are congruent and also $\angle PTQ$ is congruent to $\angle PSQ$.

Click to see why the two angles are congruent

If one wants parabola to emerge as a conic section, the intersecting plane has to be very special. Its inclination angle to the horizontal plane has to be the same as the inclination angle of any line lying on the conical surface going thru its vertex $V$, because $SP$ is on the parabola plane, which shares the same angle with the cone's side; $TP$ is on the cone, which also share the same angle with the cone's side.

$\angle PTQ=\angle PSQ$

Since the two triangles are congruent, we can make the conclusion that:

Eq. 2         $|PF| = |PT| = |PS| =$ the distance from point $P$ to the line $p$.

It means that the distance of arbitrary point $P$ of the investigated curve from point $F$ is equal to its distance from line $p$.

Therefore, the investigated curve is a parabola and line $p$ is its directrix.

There is another way to prove Sum of Distances to Foci Property.

Figure 6. Created by Nabla

See Figure 6. As the generator approaches the position needed to be parallel to the intersection plane, the point $P$ moves far away from $F$. This shows the basic property of the parabola that the line at infinity is a tangent.

Proof:

Since the segments $PF$ and $PM$ belong to tangents drawn from $P$ to the sphere, we can get:

Eq. 1         $PM = PF$.

Since the planes of the circles $k$ and $k'$ are parallel to each other and perpendicular to the section through the cone axis, and since the intersection plane is parallel to the slanting edge $VB$, the intersection $d$, of planes $E$ and $K$, is also perpendicular to the section through the cone axis.

$PN$ is perpendicular from $P$ to the line $d$. Thus,

Eq. 2         $PN = BA = PM$, or
Eq. 3         $PF = PN$.

Focus-Directrix Property

All the conic sections are different depending on a fixed point $F$ (the focus), a line $L$ (the directrix) not containing $F$ and a nonnegative real number $e$ (the eccentricity). The corresponding conic section consists of the locus of all points whose distance to $F$ equals $e$ times their distance to L. For $0 < e < 1$ we obtain an ellipse, for $e = 1$ we get a parabola, and for $e > 1$ we get a hyperbola.

Ellipse

Drop a hinge line straight off of the hat brim to a point on ellipse. It forms a constant angle with the cone wall (on which the blue hypotenuse rests) and a constant angle with the cutting plane (on which the red hypotenuse rests). The blue hypotenuse is the point's distance to focus. The red hypotenuse is thepoint's distance to directrix line. Since the triangles remain the same shape and share a leg, the ratio of the blue hypotenuse to the red hypotenuse remains constant. This constant is called the ellipse's eccentricity.

An ellipse's eccentricity is a constant number between 0 and 1. A circle's eccentricity is called zero although its directrix line is not defined.

Hyperbola

As with the ellipse, extend a perpendicular hinge from the hat brim to a point on the hyperbola. The hinge is a shared leg of 2 right triangles. The blue hypotenuse is the distance to the focus and the red hypotenuse is the distance to the directrix line. As with the ellipse, the ratio of the blue to the red hypotenuse is constant and this constant is the hyperbola's eccentricity.

A Hyperbola's eccentricity is always greater than 1.

Parabola

Dropping the hinge from the hat brim we can see the blue hypotenuse is the same as the red hypotenuse. Distance to directrix $=$ Distance to focus

and so the parabola's eccentricity is exactly 1.

For more about parabolas, see Parabola

Why It's Interesting

Other interesting Dandelin Spheres images from Hop's Gallery

Newton's Astronomy Theory

Dandelin Spheres help to prove that all astronomical objects move along conic sections around the Sun.

In A simple Cartesian treatment of planetary motion by Andrew T Hyman in 1992, the author illustrates that the two famous theorems, which we showed above, can be proved in a much simplier manner. He also proves that planets pursuing Keplerian trajectories have accelerations which conform to Neton's central $\frac{1}{R^2}$ equation. Conversely, planetary orbits must be Keplerian if Neton's central $\frac{1}{R^2}$ equation holds true.

In the book, Hyman says that Kepler introduced the first two laws in his 1609 Astronomia Nova. The third or "harmonic" las was suggested in his 1619 Harmonice Mundi and is often stated in terms of the length $a$ of the semimajor axis ($a$ is half of the orbit's greatest width).

The discovery of Kepler's laws was the greatest advance since Aristarchus deduced nineteen centuries earlier that planets circle the Sun (see Heath 1981).

Recall that ellipses are the closed curves formed by intersecting a circular cone and a plane. The ancient Greeks proved (see Heath 1981) that, everywhere along an ellipse, the distance to a point (the focus) divided by the distance to a line (the directrix) is a constant eccentricity $e$. A beautiful proof of this focus-directrix property was devised in 1822 by G P Dandelin, for both open $e \geqslant 1$ and closed $0 \leqslant e < 1$ conic sections (see Shenk 1977 or Thomas and Finney 1984).

This is a perfect example of Dandelin Spheres Theory helping to facilitate some complicated theorems.

An Interesting Math problem

The Senior Second Round of the 2006 Harmony South Africa Mathematics Olympiad contained an interesting problem in Question 20. If $AE = 3$, $DE = 5$, and $CE = 7$, then $BF$ equals:

(a) 3.6

(b) 4.0

(c) 4.2

(d) 4.5

(e) 5.0

Before reading further, the reader is encouraged to stop and first try to solve this problem.

SA Mathematics Olympiad provides the answer is (c).

By looking at the angles we see that triangles...

By looking at the angles we see that triangles $\triangle AED$ and $\triangle BFA$ are similar, as are triangles $\triangle CFB$ and $\triangle DEC$.

Therefore,

$\frac{BF}{AF} = \frac{AE}{DE}= \frac{3}{5}$

and

$\frac{BF}{CF} = \frac{CE}{DE} = \frac{7}{5}$

so

$\frac{CF}{AF} = \frac{3}{7}$.

But

$CF + AF = CE + AE = 10$

so

$AF = 3$

and

$AF = 7$.

Therefore,

$BF = \frac{BF}{AF} * AF = \frac{3}{5} * 7 = 4.2$.

However, the problem and its solution mask an interesting underlying theorem, that of the Dandelin Spheres. Since there are right angles at $F$ and $E$, we can draw two circles centered at $B$ and $D$ respectively as shown in Figure7. Assume that the two circles are of different sizes with circle $B$ smaller than circle $D$. Then draw tangents from $A$ and $C$ to circles $B$ and $D$ respectively.

This is therefore good illustrative example, as is discussed further in De Villiers, of where a 2D result can be proved much more easily by considering it as a special case of a 3D result!

Teaching Materials

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References

[1] Heath, Thomas. A History of Greek Mathematics, Clarendon Press, 1921.

[2] Kendig, Keith. Conics, Cambridge University Press, 2005.

[3] Taylor, Charles. An Introduction to the Ancient and Modern Geometry of Conics, Deighton, Bell and co., 1881.

[4] Hyman, Andrew. A Simple Cartesian Treatment of Planetary Motion, European Journal of Physics, Vol. 14, 1993.

[5] De Villiers, Michael. Further reflection on a SA Mathematics Olympiad Problem, Teaching & Learning Mathematics, No. 4, Feb 2007, page. 25-27. Retrieved from http://mysite.mweb.co.za/residents/profmd/samogeneral.pdf

[6] Wikipedia. (n.d.). Dandelin spheres. Retrieved from http://en.wikipedia.org/wiki/Dandelin_spheres

[7] Tuleja, Slavomir and Hanc, Jozef. Java applet JDandelin. 2002. Retrieved from http://www.liceomendrisio.ch/~marsan/matematica/materiale_vario/coniche/Dandelin/JDandelinEn.htm

[8] Dandelin Spheres. From MedicBD Health. Retrieved from http://www.health.medicbd.com/wiki/Dandelin_spheres#Proof_that_the_curve_has_constant_sum_of_distances_to_foci

[9] Dandelin’s Spheres - proof of conic sections focal properties. From Nabla. Retrieved from http://www.nabla.hr/Z_IntermediateAlgebraConicsFamilyOfSimilarlyShapedCurve_2.htm

[10] Conics, a family of similarly shaped curves - properties of conics. From Nabla. Retrieved from http://www.nabla.hr/Z_CollegeConicsFamilyOfSimilarlyShapedCurveProperties_2.htm

[11] David, Hollister. (n.d.). Dandelin Spheres. From Hop’s Gallery. Retrieved from http://clowder.net/hop/Dandelin/Dandelin.html

[12] Hilbert, D. and Cohn-Vossen, S. The Directrices of the Conics. Ch. 1, Appendix 2 in Geometry and the Imagination. New York: Chelsea, pp. 27-29, 1999.

• More about applications.
1. Better images for proof sections.
2. More clear step by step proof and images in section Focus-Directrix Property.

Have questions about the image or the explanations on this page?