How to Project Spherical Conics into the Plane

Transcription

1 How to Project pherical Conics into the Plane Yoichi Maeda Department of Mathematics Tokai University Japan Abstract: In this paper, we will introduce a method how to draw the orthogonal projected images of spherical conics (great circle, small circle, and conic on the sphere) in the Euclidean plane The orthogonal projected images of spherical circles are conics in the plane n the other hand, the orthogonal projected images of spherical conics are not conic but quartic in general To construct these figures with basic drawing tools, stereographic projection plays an important role The stereographic projection maps circles on the sphere to circles in the plane Using this property, we can construct the orthogonal projected images of spherical circles in the plane As for spherical conics, the famous Pascal s theorem (mystic hexagon) is essential The Pascal s theorem is also valid for spherical geometry Applying this theorem, we can also construct the orthogonal projected images of spherical conic in the plane We will realize these constructions along with the dynamic geometry software Cabri II Plus These constructions are very instructive to understand the importance of stereographic projection and also the great fun of conic 1 Introduction In this paper, we will introduce a method how to draw the orthogonal projected images of spherical conics (great circle, small circle, and conic on the sphere) in the Euclidean plane ur ordinary image of the sphere is the orthogonal projected image Under this projection, great circles of the sphere are projected to conics tangent to the boundary circle, which is the orthogonal projected image of the sphere What is the algorithm how to draw this conic passing through any two points on the sphere? This problem is the start point of our studies To solve this problem, we use an important property of stereographic projection of the sphere, that is, circle-to-circle correspondence (see [], pp74-77) To draw a conic as the orthogonal projected image, we use a circle as the stereographic projected image In the similar way, we can construct the orthogonal projected image of the small circle passing through three points We will also consider the algorithm how to draw the orthogonal projected images of the spherical conics passing through five points on the sphere This problem is solved by the spherical version of Pascal s theorem (see [1], pp ) in the Euclidean plane The orthogonal projected images of spherical conics are quartic curves in the plane and given by the locus of the stereographic projected images (see, Figure 11) For these constructions, the dynamic geometry software Cabri II Plus can be used rthogonal and tereographic Projections As mentioned in the previous section, the relation between orthogonal projection and stereographic projection is the key to solve our problem Let us first investigate this relation In the following argument, let be the unit sphere centered at the origin (0,0,0) and B be the unit disk in the XY-plane (R ) as the orthogonal projected image of from z-direction ur input is a set of orthogonal projected points (two, three, or five points) in the unit disk B, and the output in B is the orthogonal projected images of spherical conics (great circle, small circle, or conic) on the unit sphere Then the orthogonal projection ϕ : B is defined as follows: ϕ ( x, y, z) ( x, y)

2 A N B Figure 11 Quartic curve as the orthogonal projected image of a spherical ellipse Note that this map is two-to-one except for the Equator, therefore in the following argument, assume that the inputs of orthogonal projected points are regarded as the points on the northern hemisphere of n the other hands, stereographic projection of from the North Pole N(0,0,1) into the XY-plane ϕ : \N R is defined as follows: x y ϕ ( x, y, z), 1 z 1 z In this case, this map is one-to-one tereographic projection plays very important roles in the various fields of mathematics because this projection preserves angles (conformal mapping) and also has the strong property circle-to-circle correspondence We use the latter property in the following argument The map ϕ projects the northern (southern) hemisphere into the outside (inside) of B, respectively The important fact is that the Equator is fixed under both projections ϕ and ϕ ur actual constructions are done in the XY-plane, hence the maps ϕ o and ϕ o are essential The map ϕ o and also the map ϕ o ϕ ϕ : B R is written as follows: ϕ ϕ x y ϕ o ϕ ( x, y),, (1) 1± 1 x y 1± 1 x y : R B is written as follows: ( x y ϕ o ϕ x, y), () x + y + 1 x + y + 1 Geometrical constructions of these maps determined by the equations (1) and () are very simple The algorithms are as the followings (see Figures 1 and ): Algorithm 1(tereographic projected points from an orthogonal projected point) Input: orthogonal projected point P in B 1 Draw the line P, and two lines L 1 and L perpendicular to P passing through and P Let N be one of the intersections of L 1 and the unit circle Let P 1 and P be two intersections of L and the unit circle Assume that P 1 and N are on the same side of P 3 Let Q 1 be the intersection of N P 1 and P Let Q be the intersection of N P and P

3 utput: Q 1 is the stereographic projected point of P on the northern hemisphere Q is that of P on the southern hemisphere Figure 1 Construction of stereographic projected points from an orthogonal projected point Algorithm (rthogonal projected point from a stereographic projected point) Input: stereographic projected point Q in R 1 Draw the line Q, and the line L 1 perpendicular to Q passing through Let N be one of the intersections of L 1 and the unit circle Let Q 1 ( N ) be another intersection of N Q and the unit circle 3 Draw the line L perpendicular to Q passing through Q 1 Let P be the intersection of L and Q utput: P is the orthogonal projected point of Q Q Q1 L N' L1 P Figure Construction of the orthogonal projected point from a stereographic projected point 3 Great and mall Circles on ince a spherical circle on is the intersection of a certain plane and, the orthogonal projected image of the spherical circle is a conic in B n the other hand, its stereographic projected image is a circle in R Combining these facts, we can get the following algorithms (see Figures 31 and 3): Algorithm 31(Great circle passing through two points)

4 Input: two orthogonal projected points P 1 and P in B 1 Create two antipodal points P 1 * and P * of P 1 and P with respect to Create three stereographic projected points Q 1, Q, and Q 1 * of P 1, P, and P 1 * 3 Let E 1 be one of the intersections of the unit circle and the circle passing through Q 1, Q, and Q 1 * 4 Draw the conic C passing through five points P 1, P, P 1 *, P *, and E 1 utput: C is the orthogonal projected image of the great circle on Q Q1 P1 C P Q1* P1* E1 P* Figure 31 Construction of the orthogonal projected image of a great circle Algorithm 3(mall circle passing through three points) Input: three orthogonal projected points P 1, P, and P 3 in B 1 Create three stereographic points Q 1, Q, and Q 3 of P 1, P, and P 3 Take two points Q 4 and Q 5 on the circle passing through Q 1, Q, and Q 3 3 Create two orthogonal projected points P 4 and P 5 of Q 4 and Q 5 4 Draw the conic C passing through five points P 1, P, P 3, P 4, and P 5 utput: C is the orthogonal projected image of the small circle on _ Q3 Q5 Q P5 Q4 P P4 Q 1 P1 P3 C Figure 3 Construction of the orthogonal projected image of a small circle

5 Note that in Figure 31 the stereographic projected image of a great circle intersects with the unit circle at two symmetric points with respect to 4 Pascal s Theorem on First, let us review Pascal s theorem (mystic hexagon) (see [1], pp and [], pp ) and its application how to draw conics in the plane R Theorem 41(Blaise Pascal) If a hexagon is inscribed in a smooth conic, the intersections of opposite sides of the hexagon are collinear (see Figure 41) Figure 41 Pascal s mystic hexagon This theorem tells us the reason why a smooth conic is determined by five points The algorithm to draw conics is as follows: Algorithm 41(Conic in the plane) Input: five points P 1, P, P 3, P 4, and P 5 in R 1 Draw any line L 56 passing through P 5 Draw four lines L 1 P 1 P, L 3 P P 3, L 34 P 3 P 4, and L 45 P 4 P 5 3 Draw the line L 0 passing through L 1 L 45 and L 3 L 56 4 Draw the line L 61 passing through L 34 L 0 and P 1 5 Create the intersection P 6 of L 56 and L 61 6 Draw the trace C of P 6 rotating the line L 56 around the point P 5 utput: C is the conic passing through five points The line L 0 is called the Pascal line ur main intention is to extend Algorithm 41 in R to that on For this purpose, we have to go back to the definition of spherical conic Definition 41(pherical conic) A spherical conic is the intersection of a sphere and an elliptic cone centered at the center of the sphere This definition is quite different from that of ordinary conic in the plane with two foci and a string with a certain length However, the next proposition assures us that Definition 41 is appropriate Proposition 41 Let C( sin c,0, cos c ) and C ( sin c,0, cos c ) be two points on where 0 < c < π / The spherical ellipse is determined as the locus of the point P on satisfying arc

6 length(cp)+arc length(c P)a where c < a < π / Then this spherical ellipse is on the intersection of and a cone centered at the origin determined by the following equation: x y + z (41) tan a tan b where cos b cosa / cosc Remark 41 Points C( sin c,0, cos c ) and C ( sin c,0, cos c ) are foci of this spherical ellipse Note that 0 < b < a In fact, Point A( sin a,0, cos a ) is one of the end points of major axis of this spherical ellipse And Point B( 0,sin b, cos b ) is one of the end points of minor axis as in Figure 11 (see [3]) Proof Let P( x, y, z ) be a point on the spherical ellipse Then, cos CP xsin c + z cos c, sin CP 1 ( xsin c + z cos c) cos C ' P xsin c + z cos c, sin C' P 1 ( x sin c + z cos c) cos( CP + C' P) cos a implies that z cos c x sin c cos a (1 ( xsin c + z cos c) )(1 ( x sin c + z cos c) ) quaring the both sides of the above equation, cos a ( z cos c x sin c) + cos a 1 ( x sin c + z cos c), equivalently, x z + 1 sin a / sin c cos a / cos c Now, using cos b cosa / cosc and x + y + z 1, x sin a /(sin a cos a tan Consequently, we get the equation (41) z + b) cos x b + y + z, As a corollary, we obtain the following Corollary 41 For any spherical conic C on, there exist two points F 1, F on and a positive number l < π such that * * C P PF + PF l P PF + PF l { 1 } { 1 } * * { P PF PF π l} { P PF PF π l} 1 1 where F 1 * and F * are the antipodal points of F 1 and F, respectively In the next, let us show that the Pascal s theorem is valid for spherical geometry Theorem 4 If a spherical hexagon is inscribed in a spherical conic, the intersections of opposite sides of the hexagon are collinear Proof Let us consider a central projection of on a certain plane Note that the central projected image of a great circle is a line in the plane ince a spherical conic is the intersection of and an elliptic cone, the central projected image of a spherical conic is a conic section in the plane Then it is clear that the Pascal s theorem in the plane implies the Pascal s theorem on the sphere

7 Now, we are ready to construct the spherical conic in the same way as Algorithm 41 Algorithm 4(Conic on the sphere) Input: five points P 1, P, P 3, P 4 and P 5 in B 1 Create five stereographic points Q 1, Q, Q 3, Q 4 and Q 5 of P 1, P, P 3, P 4 and P 5 And create five stereographic points Q 1 *, Q *, Q 3 *, Q 4 * and Q 5 * of the antipodal points P 1 *, P *, P 3 *, P 4 * and P 5 * Draw the stereographic image circle C 56 ( {Q 5, Q 5 *}) of an arbitrary great circle passing through P 5 3 Draw four stereographic image circles C 1, C 3, C 34 and C 45 of the great circles passing through {P 1, P }, {P, P 3 }, {P 3, P 4 } and {P 4, P 5 } (C ij {Q i, Q i *, Q j, Q j *}) 4 Draw the circle C 0 passing through C 1 C 45 and C 3 C 56 5 Draw the circle C 61 passing through C 34 C 0 and Q 1 6 Create two intersections Q 6 and Q 6 * of C 56 and C 61 7 Create the orthogonal projected points P 6 and P 6 * of Q 6 and Q 6 * 8 Draw the trace C of P 6 and P 6 * rotating the circle C 56 around the point Q 5 utput: C is the spherical conic passing through five points Q6 C0 T* * T Figure 4 Construction of the orthogonal projected image of a spherical conic References [1] Berger, M (1977) Geometry II pringer-verlag Berlin Heidelberg [] Jennings, G (1994) Modern Geometry with Applications pringer-verlag New York [3] Maeda, Y (005) pherical Conic and the Fourth Parameter KMITL ci J Vol5 No1 Feb 005 (pp )