The Lemniscate of Bernoulli, without Formulas

Transcription

1 The Lemniscate of ernoulli, without Formulas rseniy V. kopyan ariv: v [math.h] 4 May 014 bstract In this paper, we give purely geometrical proofs of the well-known properties of the lemniscate of ernoulli. What is the lemniscate? polynomial lemniscate with foci F 1, F,..., F n is a locus of points such that the product of distances from to the foci is constant ( F i = const). The n-th root of this value is called the radius of the lemniscate. i=1,...,n It is clear, that lemniscate is an algebraic curve of degree (at most) n. You can see a family of lemniscates with three foci in Figure 1. Fig. 1. The lemniscate with three foci. lemniscate with two foci is called a Cassini oval. It is named after the astronomer Giovanni Domenico Cassini who studied them in The most well-known Cassini oval is the lemniscate of ernoulli which was described by Jakob ernoulli in This is a curve such that for each point of the curve, the product of distances to foci equals quarter of square of the distance between the foci (Fig. ). ernoulli considered it as a modification of an ellipse, which has the similar definition: the locus of points with the sum of distances to foci is constant. (ernoulli was not familiar with the work of Cassini). It is clear that the lemniscate of ernoulli passes through the midpoint between the foci. This point is called the juncture or double point of the lemniscate. F 1 F Fig.. F 1 F = F 1 F. 1

2 The lemniscate of ernoulli has many very interesting properties. For example, the area bounded by the lemniscate is equal to 1 F 1F. In this paper we prove some other properties, mainly using purely synthetic arguments. How to construct the lemniscate of ernoulli? There exists a very simple method to construct the lemniscate of ernoulli using the following three-bar linkage. This construction was invented by James Watt: Take two equal rods F 1 and F of the length 1 F 1 F and fixed at the points F 1 and F respectively. Let points and lie on opposite sides of the line F 1 F. The third rod cinnects the points and and its length equals F 1 F (Fig. 3). Then, during the motion of the linkage the midpoint of the rod traces the lemniscate of ernoulli with foci at F 1 and F. F 1 F Fig. 3. roof. Note that the quadrilateral F 1 F is an isosceles trapezoid (Fig. 4). Moreover, triangles F 1 and F 1 are similar, because they have the common angle and the following relation on their sides holds: F 1 = F 1 =. For the same reason, triangles F and F are similar. They have the commonangle, andtheratiosofthelengthofthesideswithendpointsat is. Therefore, we can write the following equations: F 1 = F 1 = F = F. F 1 F Fig. 4. Let us remark that in the trapezoid F 1 F angles and F are equal. Since angles F and F are equal too, we obtain F 1 = F. This implies that triangles F 1 and F are similar. Therefore, F 1 = F F 1 F = = F 1.

3 Thus, we have shown that point lies on the lemniscate of ernoulli Since the motion of the point is continuous and attains the farthest points of the lemniscate, the trajectory of is the whole lemniscate of ernoulli. M F 1 F N Fig. 5. Let be the midpoint of the segment F 1 F (double point of the lemniscate). Denote by M and N the midpoints of the segments F 1 and F 1 respectively (Fig. 5). Translate the point by the vector NF 1. Denote the new point by. bserve that triangles F 1 M and N are congruent. Moreover, the following equation holds: F 1 M = F 1 = 1 F 1. Inother words, thepointsm and lieon thecircle with center at F 1 andradius 1 F 1. Using this observation, we can obtain another elegant method to construct the lemniscate of ernoulli. F 1 F Fig. 6. Let us construct the circle with center at one of the foci and radius 1 F 1. n each secant (where and are the points of intersection of the circle and the secant) chose points and such that = = (Fig. 6). The union of all points and form the lemniscate of ernoulli with foci F 1 and F. C F 1 F Fig. 7. nother interesting way to construct lemniscate with the linkages given in Figure 7. The lengths of the segments F 1 and F 1 are equal. The point is the vertex of rods and Y of the length F 1. Denote the midpoints of these rods by and C, and join them with by the another rod of the length. In the process of rotating of point around the circle each of the points and Y generates a half of the lemniscate of ernoulli with foci F 1 and F. 3 Y

4 The lemniscate of ernoulli and equilateral hyperbola. The hyperbola is a much more well-known curve. Hyperbola with foci F 1 and F is a set of all points such that value F1 F is constant. oints F1 and F are called the foci of the hyperbola. mong all hyperbolas, we single out equilateral or rectangular hyperbolas, i. e., the set of points such that F1 F = F 1 F. The lemniscate of ernoulli is an inversion image of an equilateral hyperbola. efore proving this claim, let us recall the definition of a inversion. Definition 1. Inversion with respect to the circle with center and radius r is a transformation which maps every point in the plane to the point lying on the ray such that = r. The inversion has many interesting properties, see, for example []. mong the properties, is the following: a circle or a line will invert to either a circle or a line, depending on whether it passes through the origin. We will prove here just one simple lemma that will help us later. Lemma 1. Suppose is an orthogonal projection of the point on some line l. Then inversion image of the line l with respect to a circle with center at is the circle with diameter, where is the inversion image of the point. l 1 Fig. 8. roof. Let be any point on the line l, and let be its image (Fig. 8). Since = r and = r, we obtain that triangles and are similar. Therefore the angle is a right angle and the point lies on the circle with diameter. Note that the center 1 of this circle is the inversion of the point 1, where 1 is the point symmetric to with respect to the line l. Now, let us prove that the lemniscate of ernoulli with foci F 1 and F, is an inversion of the equilateral hyperbola with foci F 1 and F with respect to the circle with center at and radius F 1. For this proof, we will use the results we obtained in the proof of correctness of the first method to construct the lemniscate (Fig. 4). Let be the point of the intersection of the lines F 1 and F and let Q be symmetric point to with respect to the line F 1 F (Fig. 9). Note that F Q F 1 Q = F F 1 = F 1 = F 1 = F 1F. Therefore, the points and Q lie on the equilateral hyperbola with foci at F 1 and F. Now it remains to show that points and Q are the images of each other under the inversion with center at and radius F 1. First, let us show that triangles F 1 and F 1 are similar. 4

5 Q F 1 F Fig. 9. The quadrilateral F 1 is a trapezoid. Therefore F 1 + F 1 = 180. lso, we have F 1 + F 1 = 180. Since the angle F 1 is equal to the angle F 1, we obtain that F 1 and F 1 are equal to each other. Since angels F and F 1 are equal, we have that F 1 + F = 180. In other words, the quadrilateral F 1 F is inscribed. Therefore, we have F F 1 = F = F 1. The last equation provided that points and are symmetric to each other with respect to the perpendicular bisector of the segment F 1. Thus, triangles F 1 and F 1 are similar because they have two corresponding pairs of equal angles. It follows that F 1 and F 1 are equal, and we have that the point Q lies on the ray. In addition, from similarity of triangles F 1 and QF 1 (it is congruent to the triangle F 1 ), we obtain F 1 = F 1 Q Q = F 1. It means that points Q and are images of each other under the inversion with center at and radius F 1. Q F 1 F l Fig. 10. If we look at Figure 9, we can make another observation: the points and lie on the circle with centered at. It is interesting that this circle touches the lemniscate of ernoulli. roof. Suppose l is a tangent line to the hyperbola in the point Q. From Lemma 1, it follows that the image of the line l under the inversion with center at and radius F 1 5

6 is a circle ω l passing through the point. Since is the inversion image of the point Q, we see that the circle ω l touches the lemniscate at the point. From the same Lemma we conclude that the center of this circle lies on the normal line from the point to the line l. Let us show that lines and Q are symmetric to each other with respect to the line F 1 F. It follows that the point is the center of the circle ω l. F 1 S Q Fig. 11. Without loss of generality, we can assume that the equation of the hyperbola is y = 1 x. Suppose line l intersects the abscissa and the ordinate in the points R and S respectively (Fig. 11). It is well-known that the derivative of the function 1 x at the point x 0 is equal to 1 x 0. It follows that the point Q is the midpoint of the segment RS, and Q is the median of the right triangle RS. Therefore, the angles QR and QR are equal. Since angles S and QR are also equal, we obtain that the lines and RS are perpendicular, as was to be proved. α F 1 α F Fig. 1. Let usnote thefollowing: Sincethecircle ω l touches thelemniscate at thepoint, the radius of this circle is a normal (perpendicular to the tangent line) to the lemniscate at (Fig. 1). Note that the triangle is isosceles, and the lines and are symmetric with respect to the line F 1. Therefore, we can write these equations: R = = F 1. It follows that there exists the following very simple method to construct the normal to the lemniscate of ernoulli: For any point on the lemniscate, take the line forming with the line intersecting at, with the angle of intersection equal to F. This line will be a normal to the lemniscate. References [1] dir/lemniscatefernoulli dir/lemniscatefernoulli.html. [] R.. Johnson. dvanced Euclidean Geometry. Dover ublications New York, N. Y, 007. [3] J. D. Lawrence. catalog of special plane curves. Dover ublications New York, N. Y,