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Transcript
```13
13.1
A Glimpse at Elliptic Geometry
Elliptic geometry is derived from spherical geometry
Problem 13.1. This is a quite interesting interpretation of the undeﬁned terms of
incidence geometry:
A "point" is interpreted as a pair {P, Pa } of antipodal points on the unit sphere S 2 in
Euclidean three dimensional space.
A "line" is to be a great circle C on the sphere.
A "A point {P, Pa } lies on a line C" means that the great circle C contains
both antipodal points P and Pa .
Question. Which of the axioms of incidence geometry are satisﬁed, and which not?
Answer. All three axioms (I.1),(I.2) and (I.3) are satisﬁed.
Figure 13.1: A spherical line
Reason for (I.1). Given two ”points” {P, Pa } and {Q, Qa }, they determine in the
ambient Euclidean three space two diﬀerent lines P Pa and QQa , which both pass
through the origin (0, 0, 0). The two lines determine a plane , which intersects S 2
in a great circle . This great circle is the ”line” connecting the two given points.
585
Reason for (I.2). Given two ”points” {P, Pa } and {Q, Qa }, they determine the plane
uniquely, and hence the great circle from above uniquely . Hence the ”line”
connecting the two given points is unique.
Reason for (I.3). Every great circle contains inﬁnitely many points, and hence more
than one antipodal pair of points. The three points (±1, 0, 0), (0, ±1, 0) and
(0, 0, ±1) do not lie on a great circle.
Question. Which parallel property (elliptic, Euclidean, hyperbolic, or neither) does
hold?
Answer. The elliptic parallel property holds. Any two great circles intersect. Hence
there do not exist parallel lines.
13.2
The conformal model
Deﬁnition 13.1 (The stereographic model of elliptic geometry). Consider the
following model of elliptic geometry, built in the Euclidean plane as ambient underlying
reality. Throughout, we denote the open unit disk by
D = {(x, y) : x, y real, x2 + y 2 < 1}
Its boundary is the unit circle
∂D = {(x, y) : x, y real, x2 + y 2 = 1}
The center of D is denoted by O.
• The "points" are the points in the interior of the unit disk D as well as pairs of
antipodal points on its boundary ∂D.
• The "lines" are closed circular segments that pass through a pair of antipodes
on ∂D.
Problem 13.2 (Construction of an elliptic line). Construct the elliptic line through
any two given points P and Q.
The solution uses pairs of antipodes and the Lemma below.
Deﬁnition 13.2 (The antipode). For any given point P , we deﬁne the antipode Pa
−→
to be the point on the ray opposite to OP such that OP · OPa = 1. For the center O
itself, the antipodal point is the point at inﬁnity.
Lemma. If a circle passes through a pair of antipodes, it consists entirely of pairs of
antipodes. Hence a circle is an elliptic line if and only if it passes through a pair of
antipodes.
586
Proof of the Lemma. Assume that a circle C passes through a pair of antipodes P and
Pa . Let Q be a third point on the circle. We draw the chords P OPa and QO. Let Q2
be the second intersection point of the (Euclidean) line OQ with circle C. Because of
the theorem of chords (Euclid III.35), we know that
OP · OPa = OQ · OQ2
By deﬁnition of antipodes
OP · OPa = 1
and hence
OQ · OQ2 = 1
The point O lies inside the circle C, and hence Q and Q2 lie on diﬀerent sides of O, (O
lies between Q and Q2 .) This conﬁrms that the second intersection point Q2 = Qa is
the antipode of Q .
Question. Provide the drawing for the Lemma.
Figure 13.2: An elliptic line consists of pairs of antipodes
Construction 13.1 (The antipode). To construct the antipode of a given point P ,
one uses the theorems related to the Pythagorean Theorem. One erects the perpendicular
h to radius OP at point O . Let C and Ca be the intersections of h with ∂D. Next one
erects the perpendicular on CP at point C. The antipodal point Pa is the intersection
at that perpendicular with the line OP .
587
Indeed, by Thales’ theorem, the four points P, C, Pa and Ca lie on a circle. Hence
Euclid III.35 implies that OP · OPa = OC · OCa = 1.
Figure 13.3: Construction of the antipode
Question. Provide the drawing for this construction.
Construction 13.2 (Elliptic line through two given points). Let two points P
and Q be given. To ﬁnd the elliptic line through P and Q, we construct the antipode Pa
of one of the two points, say P . In the special case that P lies on ∂D, the construction
is easy, since P Pa is a diameter of D. If P lies in the interior of D, and is not the
center O itself, we proceed as explained above.
Finally, we need a circle C through the three points P, Pa and Q. Its center is the
intersection of the perpendicular bisectors of segments P Q, P Pa and QPa .
Reason for validity of the construction. Thus the circle C consists entirely of pairs of
antipodes. Hence the two circles C and ∂D intersect in the pair of antipodes A and Aa .
Hence C is an elliptic line.
Problem 13.3. Do the construction for an example. Put in—as a check—the common
chord of the two circles. It should come out to be a diameter of ∂D.
13.3
Falsehood of the exterior angle theorem
Proposition 13.1 (No Exterior Angle Theorem). The Exterior Angle Theorem
does not hold in spherical geometry. Neither does it hold in elliptic geometry.
588
Figure 13.4: Construction of an elliptic line
The following construction gives a counterexample. For simplicity, take antipodes B and
D = Ba on the equator ∂D. We connect them by two diﬀerent half great circles a and
c, lying symmetric to the south pole O. Next draw a segment through O that cuts c
at point A and a at point C, but not at a right angle. Nevertheless, you get a ﬁgure
which is point symmetric by point O. Hence the triangle ABC has the interior angle
γ = ∠ACB congruent to the nonadjacent exterior angle δ = ∠DAC.
Question. Draw the ﬁgure, as explained, in the conformal model, just the southern
hemisphere. Measure and report the three angles of your ABC. What do you observe
= CD are congruent, and segments BC ∼
Proposition 13.2 (No SAA Congruence). SAA congruence does not hold in spherical geometry. Neither does it hold in elliptic geometry.
Proof. The following construction gives an example to conﬁrm this claim. Choose any
angle γ < 45◦ . We construct two non-congruent triangles with AB = 45◦ , α = 90◦ and
γ as given.
Take two pairs of antipodes A, Aa and B, Ba on the equator with AB = 45◦ . We
chose for C any point on the diameter AOAa , thus getting a right angle α = 90◦ . Let
C2 be the point such that segment CC2 has midpoint O. Finally, one needs to get the
great semi-circles BCBa and BC2 Ba . They are point symmetric with respect to center
O.
589
Figure 13.5: The exterior angle can be congruent or smaller than a nonadjacent interior
angle.
Question (a). Draw the ﬁgure as explained into the southern hemisphere of the conformal model. Stress the triangles ACB and AC2 B with two diﬀerent colors.
Question (b). Measure and report the angles γ = ∠ACB and γ2 = ∠AC2 B.
Question (c). Which pieces of the two triangles are pairwise congruent?
Answer. Obviously the common side AB is congruent to itself. Both triangles have a
right angle at vertex A. Because of point symmetry of the entire ﬁgure by O, they are
congruent. Angles γ1 = ∠ACB ∼
= γ2 = ∠AC2 B are congruent, too.
Question (d). For which congruence theorem are these just the matched pieces? Nevertheless, the two triangles ACB and AC2 B are not congruent!
Answer. The matching pieces are those required for SAA congruence (Theorem 25 in
Hilbert’s Foundations).
Question (e). What do you observe about the two (corresponding but not congruent)
sides of triangles ACB and AC2 B, which lie across the second given angle, and were
not prescribed initially?
Answer. These are the two sides BC and BC2 . They turn out to be supplementary.
590
Figure 13.6: Two non congruent triangles matching in two angles and the side across one
of them.
Figure 13.7: SAA congruence in neutral geometry
13.4
Area of a spherical triangle
Proposition 13.3 (The area of a spherical triangle). The area of a spherical
triangle is proportional to the excess of its sum of angles over two right angles. With its
angles measured in radians, the area of a spherical ABC on a sphere of radius R is
A = (α + β + γ − π) R2
By a lune LA is denoted the region lying between two half great circle.
Question (a). Let α be the angle at vertex A between the two great half circles bounding
the lune, measured in radians. Find a formula for the area the area |LA | of this lune.
591
Figure 13.8: Another example of two non congruent triangles matching in two angles and
the side across one of them.
Answer. The area |LA | is proportional to α. For α = π, one gets a half sphere which
has area 2π R2 . Hence the area of the lune between to half great circle intersecting at
angle α is
|LA | = 2αR2
Now we want to ﬁnd the area of ABC. To get an easy drawing, I use the conformal
model. We can assume that vertices A and B lie on the equator ∂D, and vertex C on
the southern hemisphere, I mean inside disk D.
Question (b). Draw an example. I ﬁnd the case with acute angles α and β easier to
draw. Extend the side AC to the antipode Aa on the equator ∂D, and furthermore to
the antipode Ca on the northern hemisphere outside the disk D. Similarly, extend the
side BC to the antipode Ba on the equator ∂D, and furthermore to the antipode Ca
on the northern hemisphere, outside the disk D. Check whether the three points C, Ca
and O lie on a Euclidean line.
In the ﬁgure, we use three lunes:
(a) the lune LA with tips A and Aa , bounded by the extended sides ABAa and ACAa
of ABC.
592
Figure 13.9: A spherical triangle
(b) the lune LB with tips B and Ba , bounded by the extended sides BABa and BCBa
of ABC.
(c) the lune LC with tips C and Ca , bounded by the extended sides CAa Ca and CBa Ca .
The ﬁrst two lunes lie totally on the southern hemisphere (interior of disk D). Only the
third lune wraps around the equator to the northern hemisphere (exterior of disk D).
Question (c). You may color the three lunes with three diﬀerent colors. (I have used
diﬀerent textures instead only for photocopying.) What are the areas of the three lunes?
(1)
|LA | = 2α R2 , |LB | = 2β R2 , |LC | = 2α R2
We now derive two equations for areas. The two lunes LA and LB overlap in the
ABC. They cover the entire southern hemisphere except of the CAa Ba . Since a
hemisphere has area 2π R2 , one gets for the areas
(2)
|LA | + |LB | = |LA ∩ LB | + |LA ∪ LB | = |ABC| + 2π R2 − |CAa Ba |
The lune LC consists just of the two non-overlapping triangles Aa Ba Ca and CAa Ba .
Hence
(3)
|LC | = |CAa Ba | + |Aa Ba Ca |
593
Question (d). Add the two equations (2)(3) and use (1). Get the area of ABC.
|LA | + |LB | + |LC | = |ABC| + 2π R2 − |CAa Ba | + |CAa Ba | + |Aa Ba Ca |
= |ABC| + 2π R2 + |Aa Ba Ca |
= 2|ABC| + 2π R2
Because the areas of the lunes are |LA | = 2α R2 , |LB | = 2β R2 , |LC| = 2α R2 , we
ﬁnally get
|ABC| = (α + β + γ − π) R2
13.5
Does Pythagoras’ imply the parallel postulate?
A purely geometric proof would have to struggle with some obvious diﬃculties. At ﬁrst
the squares of the sides of a triangle cannot be interpreted as area of any ﬁgure in neither
spherical nor hyperbolic geometry. In these geometries, there do not exist any similar
ﬁgures other than congruent ones, and squares do not exist at all. So the squares of the
sides can only be interpreted as numbers. I am for now considering these diﬃculties as
not crucial. Here are possible approaches to address the question posed:
(1) One can look at the special Pythagorean triples of integers such that a2 + b2 = c2 .
The smallest one is the triple 32 + 42 = 52 . So a meaningful question is whether
there exist right triangles with the sides 3, 4 and 5 in either spherical or hyperbolic
geometry.
(2) Take any right triangle ABC. In spherical geometry its sides satisfy
cos c = cos a cos b
Does this imply that Pythagoras’ Theorem can never hold?
(3) Take any right triangle ABC. In hyperbolic geometry—with Gaussian curvature
K = −1—the sides of a right triangle satisfy
cosh c = cosh a cosh b
Does this imply that Pythagoras’ Theorem can never hold?
I do no know about any approach to answer at least part of these questions in the
framework of neutral geometry—where they would naturally ﬁt, similarly to Legendre’s
Theorems about the angle sum. My approach problems (2) and (3) uses substantial
tools from trigonometry and calculus.
594
13.5.1
Main results obtained by calculus
Theorem 13.1. If a spherical right triangle has the short arc as its hypothenuse c, this
side is strictly shorter than for the Euclidean ﬂat triangle with legs of the same lengths.
Hence in single elliptic geometry, the hypothenuse c of any right triangle is strictly
shorter than for the corresponding Euclidean ﬂat triangle with legs of the same lengths.
Theorem 13.2 (”The lunes of Pythagoras”). For spherical right triangles where
the hypothenuse c is the longer arc, this side may be either shorter or longer, or of equal
length as for the Euclidean ﬂat triangle with legs of the same lengths.
Assume the length of the hypothenuse is restricted to the interval c ∈ (π, 2π). The
legs satisfy a2 + b2 = c2 —as in the Euclidean case— if and only if (a, b) lies on one
curve C ⊂ E, which connects the boundary points (s∗ , 0) and (0, s∗ ) inside the quarter
circle
E = {(a, b) : 0 < a < 2π , 0 < b < 2π , a2 + b2 < 4π 2 }
Here the number s∗ ≈ 257◦ is the unique solution of tan s∗ = s∗ ∈ (π, 3π
).
2
Theorem 13.3. In hyperbolic geometry, the hypothenuse c is longer than for the Euclidean ﬂat triangle with legs of the same lengths.
13.5.2
Examples for the ”lunes of Pythagoras”
Deﬁnition 13.3 (”The lunes of Pythagoras”). The lunes of Pythagoras are right
spherical triangles, the sides a, b, c ∈ (0, 2π) of which satisfy both
(13.1)
cos c = cos a cos b
c2 = a2 + b2
and
On the pages 596 through 599, one can see several examples—obtained by solving
the equation
√
cos a2 + x2 − cos a cos x
:= g(a, x) = 0
x2
numerically for several values of a. I show the ﬂat triangle, its sides in radian measure,
together with the spherical triangle in stereographic projection, with side b on the equator (red), and side a extended through the south pole (blue). Sides c and b intersect in
both vertex A and its antipode. The lengths of the sides are given in degree measure.
595
Figure 13.10: The Pythagoras lune with a = 45◦ .
Figure 13.11: The Pythagoras lune with a = 70◦ .
596
Figure 13.12: The Pythagoras lune with a = 90◦ is highly degenerate.
Figure 13.13: The Pythagoras lune with a = 120◦ .
597
Figure 13.14: The Pythagoras lune with a = 150◦ .
Figure 13.15: The Pythagoras lune with a = 180◦ corresponds to a ﬂat right triangle with
sides 3, 4 and 5—it is degenerate, too.
598
Figure 13.16: An approximately isosceles lune.
13.6
The stereographic projection
Deﬁnition 13.4 (The stereographic projection). In Euclidean 3-space, take a
sphere S 2 of diameter 1. The stereographic projection is the central projection which
maps S 2 onto the tangent plane T to the sphere at the south pole S, with the north
pole N as center of projection. The north pole N itself is mapped to a single point ∞
at inﬁnity.
The equator is mapped to the unit circle ∂D. The southern hemisphere is mapped
to the disk D, and the northern hemisphere is mapped to the exterior R2 ∪ { ∞ } \ D
of the disk.
Main Theorem 23 (Properties of the stereographic projection).
(1) Circles are mapped to circles or Euclidean lines.
(2) Angles are preserved.
13.6.1
The stereographic projection is an inversion
To master this tricky three-dimensional problem, it is helpful to use three-dimensional
inversion, deﬁned by the straightforward generalization of deﬁnition 13.5. Thus the
preservation of circles by the stereographic projections can be derived from the preservation of circles by inversion.
599
Let B be the open interior of a three dimensional ball of radius R and center Z, and
denote its boundary sphere by ∂B.
Deﬁnition 13.5 (Inversion by a sphere). The inversion by the sphere ∂B is deﬁned
to be the mapping from the Euclidean 3-space plus one point ∞ at inﬁnity to itself,
which maps an arbitrary point P = Z, to its inverse point P —deﬁned to be the point
−→
on the ray OP such that |OP | · |OP | = R2 . Hence, especially, all the points of ∂B are
mapped to themselves. The inversion maps the origin Z to ∞, and ∞ to Z.
Proposition 13.4 (The stereographic projection as an inversion). Let ∂B be the
sphere with center Z = N at the north pole through the south pole S. Since the principal
sphere S 2 has diameter 1, the new sphere ∂B has radius 1.
Inversion by ∂B maps the sphere S 2 to the the tangent plane T at the south pole
S. The stereographic projection is the restriction of the inversion by ∂B to the principal
sphere S 2 .
Proof. Again the drawing plane is determined by the meridian S 1 through the north
pole N , south pole S and the generic point A. Let ∂B be the intersection of ∂B with
the drawing plane—which is the circle around the north pole N through the south pole
S. Inversion by ∂B maps the north pole N to inﬁnity, and the south pole S to itself.
The diameter N S is mapped to itself. The circle with diameter N S and the tangent t at
the south pole are both orthogonal to this diameter N S. Hence preservation of angles
implies that the circle with diameter N S is mapped to the tangent t.
Rotation of the entire ﬁgure around the axis N S yields a three dimensional ﬁgure,
from which one sees that inversion by ∂B maps the sphere S 2 to the the tangent plane
T at the south pole S. This mapping takes any point A ∈ S 2 along the projection ray
−−→
N A to the image point P ∈ T . Hence it is just the stereographic projection.
Direct check using triangle geometry. Take any generic point A on the median S 1 and
let P be the image of A by the stereographic projection. From the similar right triangles
N SA and N P S, one gets the proportion
|N A|
|N S|
=
= cos(∠AN S)
|N S|
|N P |
and hence
|N A| · |N P | = |N S|2 = 1
which conﬁrms that point P is the inverse image of A by a circle of radius 1 around
north pole N , as to be shown.
Reason for item (1) of the Main Theorem. We show that the image of any circle on the
sphere S 2 is a circle or line in the tangent plane T through the south pole S.
Take any circle C on the sphere S 2 . Denote its center by A. In the special case
that the center A is north pole N or south pole S, one can immediately show that C
600
Figure 13.17: The stereographic projections maps circles to circles.
is mapped to a circle. Hence we exclude this special case, and let B and C be the two
points of the circle lying on the meridian N A. Suppose the stereographic projections
maps the points A, B and C to the image points P, Q and R in the tangent plane T .
We use the plane through the north pole N , south pole S and point A as our drawing
plane. Points B, C, P, Q, R lie in this plane, too.
Let U be the circle through the points B, C and Q. Because the pairs A, P , B, Q
and C, R are inverse points by circle ∂B,
|N B| · |N Q| = |N C| · |N R| = 1
Now Euclid’s theorem of chords implies that circle U consists entirely of pairs of inverse
points. Hence all four points B, C, Q, R lies on circle U . Furthermore circle U is orthogonal to ∂B. Because of the orthogonality, inversion by the circle ∂B maps the circle U
to itself.
Rotation around the common diameter of the orthogonal circles U ⊥ ∂B produces
the two orthogonal spheres U ⊥ ∂B. Again because of the orthogonality, inversion by
the sphere ∂B— which is just the stereographic projection—maps U to itself.
Since the stereographic projection maps the sphere S 2 to the tangent plane T , it
maps the intersection S 2 ∩ U to the intersection U ∩ T . But U ∩ S 2 clearly is a circle.
Because the centers of both spheres U and S 2 lie in the drawing plane, this circle has
diameter BC, and hence U ∩ S 2 = C is the originally given circle. Its stereographic
image is U ∩ T , which is a circle in the tangent plane T , as to be shown. Again, because
the center of sphere U lies in the drawing plane, the image circle U ∩ T has diameter
QR.
601
Reason for item (2) of the Main Theorem. Let any three points A, B, C on the sphere
be mapped to the points P, Q, R on the tangent plane t. The two lines P Q and P R
have as preimages two circles ABN and ACN on the sphere S 2 .
It is enough to show that the angle ∠QP R between the two lines is congruent to the
angle between the two circles. We can measure the angle between these two circles as
the angle between their tangent at the second intersection point N .
The tangent t1 to the circle ABN at the north pole N is a line in the plane spanned
−−→
−−→
by the projection rays N A and N B. It is the parallel to line P Q through the north
pole N . Similarly, the tangent t2 to the circle ACN at the north pole N is the parallel
to line P R through the north pole N . Because parallel shift leaves angles invariant, the
angle between the lines P Q and P R is congruent to the angle between the tangents t1
and t2 , as to be shown.
13.6.2
Many congruent angles
Figure 13.18: Antipodes on the sphere are mapped to antipodes in the conformal model.
Proposition 13.5 (Antipodes).
(1) Antipodes A and Aa on S 2 are mapped to antipodal points P and Pa with respect to
the circle ∂D.
(2) Great circles (elliptic lines) are mapped to circles or Euclidean lines through a pair
of antipodes.
(3) Points on the northern and southern hemispheres of S 2 which are symmetrical to
the equatorial plan, are mapped to inverse points with respect to the circle ∂D.
602
Reason for item (1). Draw a section in the plane with points N, S and the generic point
A on the sphere S 2 . Let P be the image of A. The points O, P, Aa , Pa lie in the same
plane. Angle ∠AN Aa = 90◦ by Thales’ theorem, and angle ∠N SP = 90◦ , because the
radius is orthogonal to the tangent by Euclid III.16. The altitude theorem for the right
triangle Pa N P yields
|SP | · |SPa | = |N S|2 = 1
Hence P and Pa are antipodal points with respect to the unit circle ∂D.
Figure 13.19: Many congruent angles
Let S 1 be the meridian through point A. This is just the great circle lying in our
drawing plane. Let a be the tangent line to the circle S 1 at point A. Let R be the
intersection point of that tangent with line t = SP .
603
Let β be the congruent base angles of the isosceles triangle N OA. We claim that
triangle RAP is isosceles, too. Indeed, angle addition at vertex A shows that
β + 90◦ + ∠RAP = 180◦
and the angle sum of the triangle N P S yields
β + 90◦ + ∠RP A = 180◦
Hence the triangle RAP has congruent base angles
α = ∠RAP ∼
= ∠RP A
and hence it is isosceles.
Question. Here is a drawing of what has been explained so far. In this ﬁgure, ﬁnd and mark
with same color angle α = 90◦ − β at vertices N, P, Pa , A, Aa ﬁve times. Find and mark
with another color of your choice angle β at these vertices as often as possible. (I do
not take into account vertical angles.)
Figure 13.20: Points symmetric to the equatorial plane are mapped to inverse points with
respect to the equator circle ∂D.
Reason for item (3). Take any generic point B on the median S 1 and let Q be the image
by the stereographic projection. The antipode Ba has the image point Qa such that
|SQ| · |SQa | = 1
and the south pole S lies between Q and Qa . Rotating the antipode Ba by 180◦ around
the axis N S yields the point C, such that B and C are reﬂection images by the equatorial
604
plane. The stereographic projection maps point C to point R, which is obtained from
point Qa by a rotation of 180◦ around the south pole S. Hence
|SQ| · |SR| = |SQ| · |SQa | = 1
and the south pole S lies outside the segment QR. Clearly this conﬁrms that R is the
inversion image of Q by the equator circle ∂D.
Direct check of item (3) with triangle geometry. Take any generic point B on the median S 1 and let Q be the image by the stereographic projection. From the similar right
triangles N SB and N QS, one gets the proportion
|QS|
|BS|
=
= tan(∠BN S)
|N S|
|N B|
Now let point C be the reﬂection image of B by the equatorial plan, and R be the
stereographic image of C. From the similar right triangles N SC and N RS, one
gets the proportion
|RS|
|CS|
=
= tan(∠CN S)
|N S|
|N C|
Since segments BC and N S are parallel, the triangles N SB ∼
= SN C are congruent.
Hence |BS| = |N C| and |N B| = |SC|, and ﬁnally
|BS| |CS|
|QS| |RS|
·
=
·
=1
|N S| |N S|
|N B| |N C|
and hence
|QS| · |RS| = |N S|2 = 1
which conﬁrms that the stereographic images Q and R of the symmetric points B and
C are inverse images by the equator circle ∂D.
605
Figure 13.21: The diameter in the drawing plane is projected symmetrically.
606
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