Sunday, July 5, 2015

Spherical Geometry (Legendre 462-468)

I hope that everyone enjoyed the Fourth of July weekend. But now it's time to get back to Legendre and his description of spherical geometry.

462. The sum of the sides of any spherical polygon is less than the circumference of a great circle.

Demonstration. Let there be, for example, the pentagon ABCDE; produce the sides AB, DC, till they meet in F; since BC is less than BF + CF, the perimeter of the pentagon ABCDE is less than that of the quadrilateral AEDF. Again, produce the sides AE, FD, till they meet in G, and we shall have ED < EG + GD; consequently the perimeter of the quadrilateral AEDF is less than that of the triangle AFG; but the last is less than the circumference of a great circle (461); therefore, for a still stronger reason, the perimeter of the polygon ABCDE is less than this same circumference.

We see that here Legendre is generalizing his Proposition 461. The earlier proposition asserted that the perimeter of a spherical triangle is less than the circumference of the great circle, and now we see that this is true for the perimeter of any spherical polygon, not just the triangle.

If the sphere is the unit sphere of radius 1, then this theorem shows that the perimeter of the polygon must be less than 2pi. Actually, recall that in my last post, I mentioned a new constant -- tau -- that is equal to 2pi. So we are to see that the perimeter of the polygon must be less than tau.

Legendre's Proposition 463 isn't really a theorem, but what he calls a "scholium." Here he states how the proof of 462 is similar to that of one of his earlier theorems.

463. Scholium. This proposition is essentially the same as that of art. 357; for, if O be the center of the polygon, we can suppose at the point O a solid angle formed by the plane angles AOB, BOC, COD, &c., and the sum of these angles must be less than four right angles, which does not differ from the proposition enunciated above; but the demonstration just given is different from that of art. 357; it is supposed in each that the polygon ABCDE is convex, or that no one of the sides produced would cut the figure.

So here Legendre refers to his earlier Proposition 357, which asserts that if we have a solid angle formed by three or more plane angles, these plane angles must add up to less than 360 degrees. We know that Legendre doesn't use "degrees" -- instead, he follows the Euclidean tradition of using the right angle as a unit. If we convert to radians, a right angle is pi/2 radians. In my last post, we found another new constant to represent pi/2 -- lambda. So Proposition 357 tells us that the angles must add up to less than 4lambda, which is the same as 2pi or tau.

As it turns out, both 357 and 463 are essentially proved using induction. Recall that we use induction to prove a base case first, then we show that if the theorem holds for n, then it holds for n + 1. In algebra, the base case is typically n = 0 or n = 1, but in geometry -- where we often use induction to prove that something is true for all polygons (n-gons) -- the base case is often n = 3, the triangle. We have used induction to prove the Center of a Regular Polygon Theorem in Section 7-6 of the U of Chicago text -- technically the proof of the Polygon-Sum Theorem in Section 5-7 is also inductive. I also point out that Dr. Hung-Hsi Wu used induction to prove part of his Fundamental Theorem of Similarity, but I've since decided that the proof is too difficult for high school students and will replace it with a Dilation Postulate instead.

For Proposition 462, the base case is for n = 3 -- that the perimeter of the triangle is less than tau. We have already proved this in the previous Proposition 461. So now we must prove the inductive case -- if the perimeter of an n-gon is less than tau, so is the perimeter of an (n + 1)-gon. So we take two sides (not adjacent, but skip a side) of our (n + 1)-gon and extend them until they extend, so that we now have an n-gon. The perimeter of the n-gon is already known to be less than tau, and the perimeter of the (n + 1)-gon must be even less than that of the n-gon. This is due to the (Spherical) Triangle Inequality, as the n-gon goes along two sides of a triangle while the (n + 1)-gon goes along the third (skipped) side of the triangle. Therefore, we conclude by the Transitive Property of Inequality that the perimeter of the of the (n + 1)-gon must be less than tau. QED

So although Legendre uses a pentagon in his proof of Proposition 463, we can use an inductive proof that works automatically for any n-gon. An argument can be made that the initial case of this proof isn't actually n = 3, but is actually n = 2. In Euclidean geometry we can't have a 2-gon, but in spherical geometry it can exist. In fact, a 2-gon is actually what Legendre calls a "lunary surface" -- which we abbreviate to "lune." We see that in Proposition 461, the arcs ABD and ACD are actually the sides of a lune. The two vertices of any lune are antipodal points, and its sides are always semicircles of length pi, so the perimeter of the lune is exactly 2pi or tau. Then the inductive case tells us that the perimeter of the 3-gon is already less than that of the 2-gon, therefore less than tau.

Legendre's Proposition 357 states that the sum of the plane angles that make up a solid angle must be less than tau radians. He proves this essentially by "flattening out" the solid angle -- he takes a plane that intersects all sides of the solid angle and uses the previous Proposition 356 (which we've already proved here on the blog) to show that each plane angle of the solid angle is less than the same angle projected onto the new plane. A good way to visualize this is to imagine that the solid angle is formed at the vertex S of a pyramid -- the points A, B, C, etc., mentioned Legendre can be the vertices of the base of the pyramid, and the point O can be any point in the plane of the base -- for example, the center of the polygonal base.

I won't take the time to show the full proof of Proposition 357, but I will mention an application of this theorem. Suppose we want to figure out how many Platonic solids there are. Recall that a Platonic solid is a completely regular polyhedron -- all of its faces are congruent regular polygons. As it turns out, we can use Proposition 357 to find all of the Platonic solids.

We start with the equilateral triangle, with each angle measuring 60 degrees. Now each vertex of our Platonic solid forms a solid angle. We need at least three plane angles to form a solid angle, but there is an upper limit to how many plane angles there can be. Proposition 357 tells us that the plane angles must add up to less than 360 degrees, and since each angle is 60 degrees, there must be fewer than six of them (since 6 times 60 is 360). So there can be three, four, or five 60-degree plane angles. The Platonic solid with three 60-degree plane angles is the tetrahedron, with four is the octahedron, and with five is the icosahedron.

If we move on to squares with their 90-degree angles, we can have three 90-degree plane angles, but not four (since 4 times 90 is 360). Three 90-degree plane angles gives us the cube. Regular pentagons have 108-degree angles. Again, we can't have four of them (since 4 times 108 is more than 360), and three 108-degree angles gives us the dodecahedron. Regular hexagons have 120-degree angles, but 3 times 120 is already 360. Since each solid angle must contain at least three plane angles, we are done, since increasing the number of sides in the polygon only increases the angle. Therefore, there are only five Platonic solids -- tetrahedron, cube, octahedron, dodecahedron, and icosahedron.

I mentioned earlier that solid angles can be measured in "steradians" the same way that plane angles can be measured in radians. We define the radian measure of an angle to be the length of the arc it would subtend if it were the central angle of a unit circle. So we define the steradian measure of a solid angle to be the area of the spherical polygon it subtends on the unit sphere.

The entire circle measures tau radians, since the circumference of the unit circle is tau. Now the surface area of the unit sphere is 4pi, or 2tau, so the entire sphere is 2tau steradians. This was actually mentioned on the comments of the 360-degree Vi Hart video I linked to in my last post -- one of the commenters wanted to call the video a "tau^3" video to emphasize that the video can rotate 360 degrees in three dimensions. Another commenter responded that it would be more accurate to refer to the view as 2tau steradians.

It may be interesting to find steradian values for the solid angles of the Platonic solids, but this is quite complex and is well above high-school level math. The solid angle of a tetrahedron, for example, measures arccos(23/27) steradians. A cube is much easier to calculate -- each solid angle of a cube subtends one-eighth of a sphere, so its measure is 2tau/8 = lambda steradians, just as each plane angle of a square is lambda radians.

Let's get back to Legendre. Proposition 464 is the next major theorem listed in Legendre, so let's give the theorem and its proof.

464. If the diameter DE be drawn perpendicular to the plane of the great circle AMB, the extremities D and E of this diameter will be the poles of the circle AMB, and of every small circle FNG which is parallel to it.

Demonstration. DC, being perpendicular to the plane AMB, is perpendicular to all of the straight lines CA, CM, CB, &c., drawn through its foot in the plane (325); consequently all of the arcs DA, DM, DB, &c., are quarters of a circumference. The same may be shown with respect to the arcs EA, EM, EB, &c., whence the points D, E are each equally distant from all of the points in the circumference of the circle AMB; therefore they are poles of this circle (441).

Again, the radius DC, perpendicular to the plane AMB, is perpendicular to its parallel FN; consequently it passes through the center O of the circle FNG (456); whence, if DF, DN, DG be drawn, these oblique lines will be equally distant from the perpendicular DO, and will be equal (329). But, the chords being equal, the arcs are equal; consequently all the arcs DF, DN, DG. &c. are equal; therefore the point D is the pole of the small circle FNG, and for the same reason the point E is the other pole.

In this theorem, Legendre tells us how to find the "poles" of every circle. We've mentioned before why we use the name "poles" -- if the great circle is the Equator of the earth, then its poles are actual North and South Poles of the earth. By this theorem, the poles of any small circle parallel to the Equator -- that is, of any parallel of latitude -- are also the North and South Poles.

Notice that any small circle on the globe is parallel to some great circle -- we simply take the plane passing through the center that is parallel to the plane containing the small circle. The poles of this great circle are the endpoints of a diameter and are thus antipodal points. It then follows that the poles of any circle on the globe are antipodal points.

The proof of Proposition 464 is straightforward. It depends on two Propositions previously mentioned on the blog, which are really definitions -- 441 (definition of pole), and 325 (definition of a line perpendicular to a plane). We also have 456, which is also proved on the blog (line joining the centers of sphere and small circle is perpendicular to plane of small circle). The remaining proposition mentioned in the proof, 329, is essentially the Perpendicular Bisector Theorem.

This theorem has many corollaries -- indeed, all of the remaining propositions to be covered today are corollaries of this theorem.

465. Corollary I. Every arc DM, drawn from a point in the arc of a great circle AMB to its pole, is the fourth part of a circumference, which, for the sake of conciseness, we shall call a quadrant; and this quadrant at the same time makes a right angle with the arc AM. For the line DC being perpendicular to the plane AMC, every plane DMC, which passes through the line DC, is perpendicular to the plane AMC (351); therefore the angle of these planes, or, according to the definition, art. 442 the angle AMD is a right angle.

Here we are able to use our new constants again. On the unit sphere, the length of a "quadrant" is exactly one-fourth of the circumference tau -- that is, lambda. The propositions used in this proof are Proposition 351, which states that if a line is perpendicular to a plane, then any plane through that line is perpendicular to the plane, and Proposition 442, which defines angle of a spherical triangle.

466. Corollary II. In order to find the pole of a given arc AM, draw the indefinite arc MD perpendicular to AM, take MD equal to a quadrant, and the point D will be one of the poles of the arc MD; or, rather, draw to the two points A, M, the arcs AD, MD, perpendicular each to AM, the point of meeting D of these two arcs will be the pole required.

Using Legendre's Proposition 466, we conclude that the poles of the Prime Meridian are on the Equator (as the Equator is perpendicular to the Prime Meridian), with Longitude 90 degrees East and West (as these are one quadrant away from the Prime Meridian).

467. Corollary III. Conversely, if the distance of the point D from each of the points A, M, is equal to a quadrant, we say that the point D will be the pole of the arc AM and that, at the same time, the angles DAM, AMD, will be right angles.

As Legendre points out, Proposition 467 is the converse of Proposition 466. Our final proposition of the day is another "scholium."

468. Scholium. By means of poles, arcs may be traced upon the surface of a sphere as easily as upon a plane surface. We see that, for example, that by turning the arc DF, or any other line of the same extent, about the point D, the extremity F will describe the small circle FNG; and by turning the quadrant DFA about the point D, the extremity A will describe the arc of a great circle AM.

If the arc AM is to be produced, or if only the points A, M are given, through which this arc is to pass, we determine, in the first place, the pole D by intersection of two arcs described from the points A, M, as centers, with an extent equal to a quadrant. The pole D being found, we describe from the point D, as a center, and with the same extent, the arc AM, or the continuation of it.

If it is required to let a perpendicular from a given point P upon a given arc AM, we produce this arc to S, so that the distance PS shall be equal to a quadrant; then from the pole S and with the distance PS we describe the arc PM, which will be the perpendicular arc required.

Here Legendre describes a rotation about a center, or pole. Every great circle is a fixed line of a rotation of any magnitude centered at its poles.

Before I conclude this post, let me point out a few things. Last week was Tau Day, so we wonder, can any day be associated with our new constant lambda = pi/2? What makes pi superior to either lambda or tau in the classroom is that Pi Day is in March, while most of the days that can be associated with either lambda or tau fall in either June or July, when school is out. Just as Pi Approximation Day is on 22/7 or July 22nd, Lambda Approximation Day can be taken as half of this, or 11/7. This could be interpreted as either July 11th -- yet another summer date -- or as November 7th. This is during the school year, so if a teacher prefers lambda to pi, she can celebrate Lambda Approximation Day in the classroom in November, during the first semester.

Returning to pi now, I just checked out a new pi book from my local library: How to Bake Pi: An Edible Exploration of the Mathematics of Mathematics, by Eugenia Cheng. I will have more to say about this book as I begin reading it in my next post.

My next post will be in a few days. Recall that I posted that I live in Los Angeles County, in the state of California. In spherical geometry, my coordinates are approximately Latitude 34 North, 118 West, and I will have something special to say about these coordinates in my next post -- hopefully.

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