Notice: This material will be included in a forthcoming (summer 2000) book with the tentative title *Experiencing Geometry in Euclidean, Spherical, and Hyperbolic Spaces*. This new book will be an expanded and updated version of *Experiencing Geometry on Plane and Sphere*. This material is in draft form and may not be duplicated or quoted without the author's written permission, except for purposes of review or trying out the material with students. As always comments are welcome and will affect the final draft. Send comments to dwh2@cornell.edu.

Chapter 17

After we have found the equations [The Laws of Cosines and Sines for a Hyperbolic plane] which represent the dependence of the angles and sides of a triangle; when, finally, we have given general expressions for elements of lines, areas and volumes of solids, all else in the [Hyperbolic] Geometry is a matter of analytics, where calculations must necessarily agree with each other, and we cannot discover anything new that is not included in these first equations from which must be taken all relations of geometric magnitudes, one to another. ... We note however, that these equations become equations of spherical Trigonometry as soon as, instead of the sides *a*, *b*, *c* we put ...

*—* N. Lobachevsky, quoted in [**NE: **Greenberg]

**Problem 17.1. Circumference of a Circle**

**a.*** Find a simple formula for the circumference of a circle on a sphere in terms of its intrinsic radius and make the formula as intrinsic as possible*.

[Hint: We suggest that you make an extrinsic drawing (similar to Figure 17.1) of the circle, its intrinsic radius, its extrinsic radius, and the center of the sphere. You may well find it convenient to use trigonometric functions to express your answer. Note that the existence of trigonometric functions for right triangles follows from the properties of similar triangles that were proved in Problem **14.3**.]

**Figure 17.1. Intrinsic radius r.**

In Figure 17.1, rotating the segment of length *r*´* *(** the extrinsic radius**) through a whole revolution produces the same circumference as rotating

Even though the derivation of the formula this way will be extrinsic, it is possible, in the end, to express the circumference only in terms of intrinsic quantities. Thus, also think of the problem:

**b.*** How could our 2-dimensional bug derive this formula?*

[Hint: By looking at very small circles, the bug could certainly find uses for the trigonometric functions that they give rise to. Then the bug could discover that the geodesics are actually (intrinsic) circles, but circles which do not have the same trigonometric properties as very small circles. And then what? Use your experience from Chapter 12.]

Using the expressions of trigonometric and hyperbolic functions in terms of infinite series, it is proved (in [**NE: **Greenberg], page 337) that

**Theorem 17.1. ***In a hyperbolic plane of radius *1*, a circle with intrinsic radius r has circumference c equal *

*c = *2p sinh(*r*).

**c. ***Use the theorem to show that on a hyperbolic plane of radius r, a circle with intrinsic radius r has circumference c equal*

*c = *2p *R* sinh(*r/**R*)*.*

[Hint: When going from a hyperbolic plane of radius 1 to a hyperbolic plane of radius *r*, all lengths scale by a factor of *r*. *Why*?]

The formula in Part **c** should look very much like your formula for Part **a** (possibly with some algebraic manipulations). This is precisely what Lobachevsky was talking about in the quote at the beginning of this chapter. Check in out in:

**d.** *Show that, if you replace R by iR in the formula of Part ***c**, *then you will get the formula in Part ***a**.

[Hint: Look up the definition of *sinh* (hyperbolic sine)* *and express it as a Taylor series.]

*Problem 17.2. Law of Cosines*

If we know two sides and the included angle of a (small) triangle, then according to SAS the third side is determined. If we know the lengths of the two sides and the measure of the included angle, how can we find the length of the third side? The various formulas that gives this length are called the ** Law of Cosines**.

**a.*** Find a law of cosines for triangles in the plane*.

**Figure 17.2. Law of Cosines.**

You have learned in school (but perhaps forgotten) the ** Law of Cosines** on the plane:

** ( a + b )^{2} = a^{2} + b^{2} + 2ab c^{2} = a^{2} + b^{2} + 2ab(-cos (**

** c^{2} = a^{2} + b^{2}**

**Figure 17.3. Three related geometric proofs. ^{1}
**

**b.** *Find a law of cosines for small triangles on a sphere with radius R*.

**Figure 17.4 . Radian measure of lengths.**

** radian measure of the arc** = (length of the arc)/

**Pause, explore, and write about this problem before you read further.**

**cos c = cos a cos b + sin a sin b cos q.
**

**(This is not the only such formula.)
**

**For a right triangle (q = p/2) the above formula becomes
**

**which can be considered as the spherical equivalent of the Pythagorean Theorem.
**

**Theorem 17.2. ***A Law of Cosines for triangles on a hyperbolic plane with radius R is*

**cosh c/R = cosh a/R cosh b/R + sinh a/R sinh b/R cos q.
**

** Closely related to the Law of Cosines is the Law of Sines.
**

** Figure 17.6. Standard proof of Law of Sines on plane.**

**b. ***What is an analogous property on the sphere*?

**Pause, explore, and write about this problem before you go to the next page.**

**When we measure the sides in radians, on the sphere the Law of Sines becomes:
**

**(sin a)/sin a = (sin b)/sin b = (sin c)/sin g.
**

**For a right triangle this becomes:
**

**Figure 17.7. Law of Sines for right triangles on a sphere.**

*Two distinct point-pairs determine a unique great circle*(*geodesic*).*Two distinct great circles determine a unique point-pair.**The center of a circle is a single point-pair.**SAS, ASA, SSS, AAA are true for all triangles not containing any point-pairs.*

(the point-pair that is the intrinsic center of the great circle).*The dual*^{2}*of a great circle is its***poles***The dual of a point-pair is its*(the great circle whose center is the point-pair).**equator**

*If the point-pair**P**is on the great circle*.**l**, then the dual of**l**is on the dual of**P****Figure 17.8***. P*is on*l*implies the dual of*l*is on the dual of*P.*If

*k*is another great circle through*P,*then notice that the dual of*k*is also on the dual of*P*. Since an angle can be viewed as a collection of lines (great circles) emanating from a point, the dual of this angle is a collection of point-pairs lying on the dual of the angle's vertex. And, vice versa, the dual of the points on a segment of a great circle are great circles emanating from a point-pair which is the dual of the original great circle.*Before going on be sure to understand this relationship between an angle and its dual.*Draw pictures. Make models.**Problem 17.4.***The Dual of a Small Triangle**The dual of the small triangle**ABC*is the small triangle D*A*B*C**, where*A**is that pole of the great circle of*BC*which is on the same side of*BC*as the vertex*A*, similarly for*B**and*C**. See Figure 17.9.**a.***Find the relationship between the sizes of the angles and sides of a triangle and the corresponding sides and angles of its dual.***b.***Is there a triangle which is its own dual?***Figure 17.9. The dual of a small triangle.*****Problem 17.5.***Trigonometry with Congruences***a.***Find the dual of the Law of Cosines on the sphere.*There is an analogous dual Law of Cosines for a hyperbolic plane which is proved in the same books, cited above, that prove the hyperbolic Law of Cosines.

**b.***For each of ASA, RLH, SSS, AAA, if you know the measures of the given sides and angles, how can you find the measures of the sides and angles that are not given*?*Do this for both spheres and hyperbolic planes.*[Hint: Use Part

**a**and the formulas from Problems**17.1**and**17.2**.]*Duality on the Projective Plane*The gnomic projection,

, (Problem*g***16.1**) allows us to transfer the above duality on the sphere to a duality on the plane. If*P*is a point on the plane, then there is a point*Q*on the sphere such that(*g**Q*) =*P*. The dual of*Q*is a great circle*l*on the sphere. If*Q*is not the South Pole, then half of*l*is in the Southern Hemisphere and its projection onto the plane,(*g**l*)*,*is a line which we can call the. This defines a dual for every point on the plane except for the point where the South Pole of the sphere rests. See Figure 17.10. It is convenient to call this point the origin,*dual of P**O*, of the plane.Note that

*O*is the image of*S*, the South Pole, and that the dual of*S*is the equator which is projected by*G*to infinity on the plane. Thus we define the dual of*O*to be the. If*line at infinity**l*is any line in the plane, then it is the image of a great circle on the sphere which intersects the equator in a point-pair. The image of this point-pair is considered to be a*single*point at infinity at the "end" of the line*l*, the*same*point at both ends. The plane with the line at infinity attached is called the.*projective plane***Figure 17.10. Duality on the plane.****Problem 17.6.***Properties on the Projective Plane**Check that the following hold on the projective plane:***a.***Two points determine a unique line*.**b.***Two parallel lines share the same point at infinity*.**c.***Two lines determine a unique point*.**d.***If a point is on a line, then the dual of the line is a point which is on the dual of the original point*.**e.***If C is the circle with center at the origin and with radius the same as the radius of the sphere, then the dual of a point on C is a line tangent to C*.*Perspective Drawings and Vision*Look at the following perspective drawing:

**Figure 17.11. Perspective drawing.**Present day theories of projective geometry got their start from Euclid's

*Optics*[**AT**: Euclid] and later from the theories of perspective developed during the Renaissance by artists who studied the geometry inherent in perspective drawings. The "vanishing point" where the lines which are parallel on the house intersect on the horizon of the drawing is an image on the drawing of the point at infinity on these parallel lines.One way to visualize this is to imagine yourself at the center of a transparent sphere looking out at the world. If you look at two parallel straight lines and trace these lines on the sphere, you will be tracing segments of two great circles. If you followed this tracing indefinitely to the ends of the straight lines, then you would have a tracing of two half great circles on the sphere intersecting in their endpoints. These endpoints of the semicircles are the images of the point at infinity that is the common intersection of the two parallel lines. If you now use a gnomic projection to project this onto a plane (e.g., the artist's canvas), then you will obtain two straight line segments intersecting at one of their endpoints as in the drawing above.

^{1}I first saw the idea for these pictures in the marvelous book [**G**: Valens].