# «1. Introduction Mathematicians and philosophers of Ancient Greece studied the problem of trisecting a general angle, doubling the cube, and squaring ...»

## GEOMETRIC CONSTRUCTIONS WITH ELLIPSES

## ALISKA GIBBINS AND LAWRENCE SMOLINSKY

Abstract. Geometric construction with a straight edge, compass, and other curves and devices was a major force in the development of mathematics. In this paper the authors examine constructions with ellipses. Trisections using hyperbolas were known

to Papus and a trisection construction using a parabola was found

by Descartes. The authors give a trisection using ellipses.

1. Introduction Mathematicians and philosophers of Ancient Greece studied the problem of trisecting a general angle, doubling the cube, and squaring the circle. They tried to accomplish these constructions using only a straight edge and compass. While these methods were unsuccessful, they also examined allowing other constructions, devices, and curves.

For example, Archimedes is usually credited with a mechanical device to trisect angles and with a construction using the spiral of Archimedes.

Pappus wrote of trisection constructions by Apollonius using conics, and he gave two constructions with hyperbolas. Menaechmus, the discoverer of conic sections, is supposed to have made his discovery while working on the problem of doubling the cube. A collection of these constructions are available on the University of St. Andrew’s website [1] There were ancient solutions to the problems of trisecting a general angle and doubling the cube, using a hyperbola and parabola. In this paper we examine constructions using ellipses. Among the results we show is that one can trisect a general angle and double the cube using ellipses.

Another trisection construction is due to Ren´ Descartes in his 1637 e La Geometrie. This construction uses a parabola and a circle relying on the the triple angle formula [2]. It is similar in spirit to the trisection via ellipses given in this paper. A version of Descartes’s trisection is shown in Figure 1. The construction uses the parabola deﬁned by y = 2x2 and the circle through the origin with center (1/2 cos(θ), 1). Solving Aliska Gibbins is a Tulane University undergraduate. She was a visiting student at Louisiana State University displaced by Hurricane Katrina when this work was done.

## 2 ALISKA GIBBINS AND LAWRENCE SMOLINSKY

2.5 1.5 0.5-1 1

-0.5 0.5 Figure 1. Descartes’s trisection of θ = π.

for the x-coordinates of the points of intersection gives the equation x(4x3 − 3x − cos(θ)) = 0 or x(x − cos(θ))(x − cos(θ + 2π/3))(x − cos(θ + 4π/3)), since a triple angle formula is cos(3θ) = 4cos3 (θ) − 3cos(θ).

The trisection for π/3 is shown in Figure 1. There are four distinct points of intersection although two are very close together.

Recent work by Carlos R. Videla explored the concept of constructability when all conics are allowed. Extending straight edge and compass constructions with all the conics, gives Videla’s notion of conically constructible. Videla allows the construction of a conic when the focus, directrix, and eccentricity are constructible. The conicallyconstructible numbers may be obtained by parabolas and hyperbolas alone. These conic constructions include ancient constructions of Papus and Menaechmus. Videla made the case that allowing a noncircular conic to be constructed if its directrix, focus, and eccentricity are constructible is consistent with the constructions of the Ancient Greeks.

Ellipses are treated as extraneous as all the constructions can be accomplished with parabolas and hyperbolas. Among the results of this paper, is that all the conic constructions can be accomplished with ellipses alone.

We propose a mechanical device that will allow us to draw ellipses.

The device consists of pins and string and allows the construction of an ellipse given the foci and a point on the ellipse. This approach is one way to construct ellipses. Allowing the construction of an ellipse given its directrix, focus, and eccentricity is another. These two approaches will be shown to allow the same constructions.

modern mathematics. We think of the plane as the Cartesian plane with coordinates. Start with an initial set of points P in the plane.

The initial set of points should include (0, 0) and (1, 0). The set of points one may derive using only a straight edge and compass will be called classically constructible points derived from P. If the set of initial points P is clear or if P = {(0, 0), (1, 0)}, then we suppress the reference to P. The set of all numbers that arise as the ordinate or abscissa of classically constructible points is the set of classically constructible numbers. Consider, for example, the question of whether one can trisect an angle of measure θ. If A = (cos(θ), sin(θ)) is in the original set of points P, then the angle determined by ∠ABC where B = (0, 0) and C = (1, 0) has measure θ. The angle can be trisected if (cos(θ/3) is a classically constructible number from P because then one may construct the point A = (cos(θ/3), sin(θ/3)) and ∠A BC.

We recall some of the facts about classically constructible numbers taking some of the background notions from Hungerford [4]. It is one of the founding observations that starting with an initial set of points P the classically constructible numbers form a ﬁeld, i.e., using straight edge and compass constructions one may start with two numbers and construct the sum, diﬀerence, product and quotient. Furthermore, if (x, y) is a classically constructible point, it is a simple exercise to show that (y, x) is a classically constructible point. It is useful to introduce the notion of the plane of a ﬁeld. If F is a subﬁeld of the real numbers R, then the plane of F is the subset of R2 consisting of all (x, y) with x, y ∈ F. If P and Q are distinct points in the plane of F then the line determined by P and Q is a line in F. Similarly, if P and Q are distinct points in F then the circle with center P and containing Q is a circle in F. It is a straightforward calculation that the intersection points of two lines in F are points in the plane of F. Furthermore if a circle in F is intersected with either a line in F or a circle in F, then the intersection points are in the plane of F (r), where r is the square root of an element of F [4].

Suppose one starts with an initial set of points P. The coordinates of these points generate a subﬁeld K of R. Since P contains (0, 0) and (1, 0), K contains the rational numbers Q. The ﬁeld of classically constructible numbers derived from P can be completely characterized.

It is the smallest subﬁeld of R containing K in which every positive number has a square root. We will also call this ﬁeld the ﬁeld of classically constructible numbers derived from K.

## 4 ALISKA GIBBINS AND LAWRENCE SMOLINSKY

3. Elliptic Constructions Consider the following description of a mechanical device that will allow the construction of an ellipse when one has the foci and a point on the ellipse accessible. The device consists of a length of string with a clip, which may be clipped to form a loop, and some pins.

(1) Given three points, one may insert pins in the three points, tighten the string around the pins, and remove one pin. Keep the string taut and use a pen to draw the curve around the two pins. The result is a curve whose distance from any point on the curve to one of the pins plus the distance from the same point on the curve to the other pin is constant—an ellipse with the pins located at the foci, which contains the third point.

** Remark 3.1.**

If an ellipse is in standard position then its equation can be given as Ax2 + Cy 2 + F = 0, √ where C A 0. Let a = F, b = F and c = a2 − b2. Then A C the foci of the ellipse are at (±c, 0) and (a, 0) is a point on the ellipse.

The tightly stretched string wraps around the foci and reaches to a point on the ellipse. It has length 2c + 2a. The directrix is given by x = ac and the eccentricty e = a.c (2) Given two points, one may insert pins in the two points, tighten the string around the pins, remove one pin, and use a pen to draw the circle with center one point and the other on the circle.

In addition we may use the straight edge.

(3) Given two points, one may draw the line through the two points.

The operations 2 and 3 have the same result as the use of a compass and straight edge. Which points can be reached by constructions with a straight edge and pins and string? We may obtain the constructible points as follows. Start with a set of points P = P0. Perform every geometric construction on the set P0 allowed by 1, 2, and 3. Adjoin to P0 all intersections of the ellipses, circles, and lines to obtain a set of points P1. We can repeat the process an arbitrary number of times.

Once Pi is constructed, perform every geometric construction on the set Pi allowed by 1, 2, and 3. We now adjoin to Pi all intersections of the ellipses, circles, and lines to obtain a set of points Pi+1. The elliptically constructible points derived from P is the union of the Pi ’s.

We will use the term constructible to mean elliptically constructible and use the term classically constructible for straight edge and compass constructions.

## GEOMETRIC CONSTRUCTIONS WITH ELLIPSES 5

We call the set of numbers that are the ordinate or abscissa of points obtainable using a straightedge and pins and string, elliptically constructible. Every straightedge and compass construction is included in construction with a straightedge and pins and string. Hence given a set of points that includes (0, 0) and (1, 0), the elliptically constructible numbers will be a ﬁeld.Analogous to the previous deﬁnitions for classical constructions is the following. Suppose F is a subﬁeld of the real numbers R. If O, P and Q are distinct points in the plane of F, then the ellipse containing O with foci P and Q is an ellipse in F.

** Lemma 3.2.**

Suppose that F is a subﬁeld of R in which every positive number has a square root. If cos (θ) ∈ F, then rotation of the plane by θ induces a bijection on the plane of F. If (r, s) is in the plane of F, then translation of the Cartesian plane by (r, s) induces a bijection on the plane of F.

Proof. Note that if cos (θ) is in F, then sin (θ) is in F since sin2 θ = 1− cos2 θ. Let f be the rotation of the plane by an angle θ, then f (x, y) = (x cos (θ) − y sin (θ), x sin (θ) + y cos (θ). If (x, y) is in the plane of F, then the image is in the plane of F. The inverse transformation of f is rotation by −θ, therefore f is a bijection.

If g is translation by (r, s), then g(x, y) = (x + r, y + s) is a bijection of the plane of F with inverse translation by (−r, −s).

** Lemma 3.3.**

Suppose that F is a subﬁeld of R in which every positive number has a square root. Suppose an ellipse is described by the equation ax2 + bxy + cy 2 + dx + ey + f = 0 with a, b, c, d, e, f ∈ F.

If cos (θ) ∈ F, then the ellipse rotated about the origin by θ can be expressed by an equation with coeﬃcients in F.

Proof. Let θ be the angle of rotation. The transformed ellipse is described by the equation a x2 + b xy + c y 2 + d x + e y + f = 0, where

**the coeﬃcients are:**

a cos2 (θ) + b sin (θ) cos (θ) + c sin2 (θ), a = 2(c − a) sin (θ) cos (θ) + b (cos2 (θ) − sin2 (θ), b = a sin2 (θ) − b sin (θ) cos (θ) + c cos2 (θ), c = d = d cos (θ) + e sin (θ), e = e cos (θ) − d sin (θ), f = f, which are all ﬁeld operations of elements of F. Therefore the coeﬃcients are in F.

## 6 ALISKA GIBBINS AND LAWRENCE SMOLINSKY

Proposition 3.4. Suppose that F is a subﬁeld of R in which every positive number has a square root. Consider an ellipse E described by the equation ax2 + bxy + y 2 + dx + ey + f = 0. The ellipse E is in F if and only if a, b, d, e, and f are in F. If E is in F, then it may also be rotated and translated to an ellipse in F in standard position.Proof. Assume the ellipse is in F. Let the points (fx, fy ) and (gx, gy ) be the focci and (x0, y0 ) be a point on the ellipse. Let d = (fx − x0 )2 + (fy − y0 )2 + (gx − x0 )2 + (gy − y0 )2 which is an element of F. The coordinates of a point on the ellipse will satisfy the equation (fx − x)2 + (fy − y)2 + (gx − x)2 + (gy − y)2 = d This equation is equivalent to Ax2 + Bxy + Cy 2 + Dx + Ey + F = 0,