Today we return to our study of spherical geometry. The next two pages of Legendre's text include nine proposition -- the first six being corollaries of the last theorem on the previous page. Recall that Proposition 452 tells us that the intersection of a sphere and a plane is a circle -- thereby justifying the names "great circle" and "small circle" for such intersections.

453. Corollary I. If the cutting plane pass through the center of the sphere, the radius of the section will be the radius of the sphere; therefore all great circles are equal to each other.

454. Corollary II. Two great circles always bisect each other; for the common intersection, passing through the center, is a diameter.

455. Corollary III. Every great circle bisects the sphere and its surface; for if, having separated the two hemispheres from each other, we apply the base of one to that of the other, turning the convexities the same way, the two surfaces will coincide with each other; if they did not, there would be points in these surfaces unequally distant from the center.

456. Corollary IV. The center of a small circle and that of a sphere are in the same straight line perpendicular to the plane of the small circle.

457. Corollary V. Small circles are less according to their distance from the center of the sphere; for, the greater the distance

*CO*, the smaller the chord

*AB*, the diameter of the small circle

*AMB*.

458. Corollary VI. Through two points on the surface a given arc of a great circle may be described; for the two given points and the center of the sphere determine the position of a plane. If, however, the two given points be the extremities of a diameter, these two points and the center would be in a straight line, and any number of great circles might be made to pass through the two given points.

I won't discuss the proofs of all of these propositions in detail -- these are all

*corollaries*, so that their proofs follow directly from that of the main theorem. But I do wish to comment on some of these:

-- Corollary V mentions that the farther a chord is from the center of a circle, the smaller it is. I point out that this is not proved in the U of Chicago text -- but is apparent simply by drawing various chords on a circle. A full proof would follow from SAS Inequality -- also known as the Hinge Theorem -- since drawing a chord farther from the center decreases its central angle.

-- Corollary III tells us that every great circle divides a sphere into two congruent hemispheres. Notice that Legendre proves this by finding an

*isometry*mapping one to the other. This isometry turns out to be a reflection, with the mirror being the plane containing the great circle. Recall that while in 2D mirrors are lines, in 3D mirrors are planes. The proof is indirect -- had the reflection image of one hemisphere not been the other hemisphere, there would exist points whose distances from the center are different, since reflections preserve distance. This would contradict the definition of a sphere as the locus of all points the

*same*distance from the center.

-- Corollary VI mentions that through three noncollinear points, there is exactly one plane. This is part (f) of the Point-Line-Plane Postulate, mentioned in Section 9-1 of the U of Chicago text. Notice that Legendre mentions a special case -- if the two points are the endpoints of a diameter. These two points have a special name -- "antipodes." So the proposition tells us that through any two points there is exactly one arc of a great circle -- unless these two points are antipodal, in which case there are infinitely many great circle arcs. As it turns out, the existence of a great circle passing through two points is of key importance -- and we'll find out why when we look at the next proposition.

459. In any spherical triangle

*ABC*, either side is less than the sum of the other two.

Notice that if we were to remove the word "spherical," this statement would remain true for ordinary triangles in the Euclidean plane. This theorem is known as the Triangle Inequality. Since Proposition 459 is the analog of the Triangle Inequality for spherical triangles, we might as well call this theorem the "Spherical Triangle Inequality." Here is how Legendre proves the Spherical Triangle Inequality:

Demonstration. Let

*O*be the center of the sphere, and let the radii

*OA*,

*OB*,

*OC*be drawn. If the planes

*AOB*,

*AOC*,

*COB*be drawn, these planes will form at the point

*O*a solid angle, and the angles

*AOB*,

*AOC*,

*COB*will have for their measure the sides

*AB*,

*AC*,

*BC*of the spherical triangle

*ABC*(123). But each of the three plane angles, which form the solid angle, is less than the sum of the two others (356); therefore either side of the triangle

*ABC*is less than the sum of the other two. QED

This proof is a bit difficult, as it depends on two other Propositions, 123 and 356. (I'm sorry, but I can't get the Jackson Five out of my mind, seeing "ABC" and "123" written so close together.)

Of these two theorems previously proved in Legendre, 123 is much easier. Recall that Legendre, like Euclid, doesn't use degrees to measure angles. Most of the time, Legendre uses the right angle as a unit, so "two right angles" means 180 degrees. But when the angle is the central angle of a circle, Legendre proves, in Proposition 123, that the measure of the central angle is proportional to the measure of its arc. (This is implied, but never proved, in Section 8-8 of the U of Chicago.)

So Legendre states that we can "define" the angle measure to be that arc length. Indeed, we still do this today -- if the circle has radius 1, we define the

*radian measure*of the central angle to be the length of its associated arc. I will have more to say about radians in a subsequent post.

But now we reach Proposition 356, the more difficult of the two used to prove 459. Here is the statement of this proposition:

356. If a solid angle is formed by three plane angles, the sum of either of these plane angles will be greater than the third.

We need to define a few terms here. A

*solid angle*is formed by the intersection of planes in the same way that an ordinary plane angle is formed by the intersection of lines. The easiest way to consider a solid angle is to look at a corner of the room. The ceiling and two walls form a solid angle.

Now solid angles can be measured just as plane angle can. In fact, since the solid angle that we want has its vertex at the center of a sphere, we can define the measure of the solid angle to be the surface area of the spherical triangle on the unit sphere in the same way that the radian measure of a plane angle is defined as an arc length. This is known as

*steradian*measure.

But Proposition 356 isn't concerned with measure of the solid angle itself. Instead, we look only at the measures of the three angles which form the solid angle. In the case of the corner of the room, each of these three angles is obviously a right angle. So the statement of the theorem is true in this case, since 90 + 90 is clearly greater than 90.

To prove Proposition 356, Legendre states that without loss of generality, we can call

*S*the vertex of the solid angle, and let

*ASB*,

*ASC*, and

*BSC*be the three angles forming the solid angle, with

*ASB*the largest of the three angles. So the statement to be proved is

*ASB*<

*ASC*+

*BSC*. We can now convert Legendre's proof into a two-column proof:

Given: Solid angle

*S*formed by plane angles

*ASB*,

*ASC*,

*BSC*

*ASB*>

*ASC*,

*ASB*>

*BSC*

Prove:

*ASB*<

*ASC*+

*BSC*

*Proof:*

Statements Reasons

1. Solid angle

*S*, etc. 1. Given

2.

*D*on

*AB*such that

*SC*=

*SD*, 2. Protractor/Ruler Postulates

Angle

*BSD*=

*BSC*

3.

*BS*=

*BS*3. Reflexive Property of Congruence

4. Triangle

*BSD*,

*BSC*congruent 4. SAS Congruence

5.

*BD*=

*BC*5. CPCTC

6.

*AB*<

*AC*+

*BC*6. Triangle Inequality

7.

*AD*+

*DB*<

*AC*+

*BC*7. Segment Addition (Betweenness)

8.

*AD*<

*AC*8. Subtraction Property of Inequality (7 minus 5)

9.

*AS*=

*AS*9. Reflexive Property of Congruence

10. Angle

*ASD*<

*ASC*10. SSS Inequality (Converse Hinge, 2, 9, 8)

11.

*ASD*+

*BSD*<

*ASC*+

*BSC*11. Addition Property of Inequality (10 plus 2)

12.

*ASB*<

*ASC*+

*BSC*12. Angle Addition

We see that this proof uses the SSS Inequality, a theorem that doesn't appear in the U of Chicago, but does appear in other texts. In some ways, it's the converse of the SAS Inequality or Hinge Theorem -- if two triangles have two pairs of congruent sides but the third pairs are different, the angle opposite the longer side is larger than the angle opposite the shorter side. It's proved very much the same way that the Unequal Angles Theorem is proved from its converse the Unequal Sides Theorem -- we use an indirect proof and assume that the angle opposite the longer side is not larger. But it can't be smaller for that would contradict the forward SAS Inequality, and the angles can't be congruent for that would contradict SAS Congruence.

We also notice that this proof uses the usual Euclidean Triangle Inequality. So this proof combined with that of Proposition 459 essentially converts the Plane Triangle Inequality into the new Spherical Triangle Inequality.

Now here's the statement that I definitely want to emphasize here -- Proposition 460:

460. The shortest way from one point to another on the surface of a sphere is the arc of a great circle which joins the two given points.

So just as the shortest distance between two points on the Euclidean plane is a straight line, the shortest distance between two points on the globe is the arc of a great circle. Fortunately, Proposition 458 guarantees that such a great circle path always exists. As it turns out, just as the shortest distance on the plane follows from the Plane Triangle Inequality, the shortest distance on the sphere follows from the Spherical Triangle Inequality:

Demonstration: Let

*ANB*be the arc of a great circle which joins the given points

*A*and

*B*, and let there be without this arc, if it be possible, a point

*M*of the shortest line between

*A*and

*B*. Through the point

*M*draw the arcs of great circles

*MA*,

*MB*, and take

*BN*=

*MB*.

According to the previous theorem, the arc

*ANB*is less than

*AM*+

*MB*; taking from one

*BN*, and the other its equal

*BM*, we shall have

*AN*<

*AM*. Now the distance from

*B*to

*M*, whether it be the same as the arc

*BM*, or any other line, is equal to the distance from

*B*to

*N*; for by supposing the plane of the great circle

*BM*to turn about the diameter passing through

*B*, the point

*M*may be reduced to the point

*N*, and then the shortest line from

*M*to

*B*, whatever it may be, is the same as that from

*N*to

*B*; consequently the two ways from

*A*to

*B*, one through

*M*and the other through

*N*, have the part from

*M*to

*B*equal to that from

*N*to

*B*. But the first way is, by hypothesis, the shortest; consequently the distance from

*A*to

*M*is less than the distance from

*A*to

*N*, which is absurd, since the arc

*AM*is greater than

*AN*; whence no point of the shortest line between

*A*and

*B*can be without the arc

*ANB*; therefore this line is the shortest that can be drawn between its extremities. QED

We notice that in this above proof, we can simply change "sphere" to "plane" and "arc (of a great circle)" to "segment" and, using the Plane Triangle Inequality, we obtain a proof that the shortest distance between two points is a straight line. Notice that this proof mentions another isometry -- we

*rotated*

*M*about the point

*B*to obtain its image

*N*.

Now we can see why great circles are so important in spherical geometry. The reason that the shortest distance between two points is the great circle path is that this proof depends on the Spherical Triangle Inequality, and the sides of spherical triangles are by definition great circle arcs. In turn, the sides are defined to be great circle arcs because that definition makes the proof of Proposition 459 work. We ultimately get back to Proposition 123, where we needed the centers of the circles to share the same center -- the center of the sphere -- so that we can identify angle measure with arc length (our little "radian" trick above). So this entire sequence of proofs must go back to great circle arcs. They are the shortest distance between two points on the sphere -- often called a "geodesic."

The idea that great circles are geodesics is one of my favorite spherical lessons. Many of the students whom I tutor will be traveling to South Korea over the summer. The flight over the Pacific Ocean will take them very close to Alaska. And so we wonder, why does the flight from California to Korea go anywhere near Alaska? Wouldn't it make more sense for the plane to cross the Pacific Ocean much closer to, say, Hawaii, than Alaska?

The reason for the Alaska flight is that this route is actually

*shorter*than the Hawaii route! The shortest distance between two points is a great circle, and the great circle passing through California and Korea is actually much closer to Alaska than to Hawaii. The following link is called the Great Circle Mapper:

http://www.gcmap.com/

I asked the Great Circle Mapper to draw the flight path from LAX here in Southern California to ICN in Korea. This is what it drew:

http://www.gcmap.com/mapui?P=LAX-ICN

And we see the red curve passing through the Aleutian Islands off the coast of Alaska. We notice that Seoul isn't that far to the north -- it's about the same latitude as San Francisco -- yet the shortest flight path passes near Alaska.

This frequently occurs on flights between different continents. There is a flight going from LAX to the Dubai, the capital of the United Arab Emirates. Despite Dubai being closer to the Equator than Los Angeles is, the shortest flight path to Dubai is actually very close to the North Pole!

http://www.gcmap.com/mapui?P=LAX-DXB

If we were to start from San Francisco or Seattle, we'd fly even closer to the North Pole. Notice that it's actually

*longitude*, not latitude, which determines how close to the Pole a flight will be. A full meridian is only half of a great circle -- the other half is the meridian 180 degrees apart. London, England, is right on the Prime Meridian (because of Greenwich and all that), and Fiji is close to 180 degrees longitude. (This would be the International Date Line, except that the line is bent in order to avoid passing through the islands.) So we expect a flight from London to Fiji to pass almost directly over the North Pole.

http://www.gcmap.com/mapui?P=LHR-NAN

There aren't many flights passing near the South Pole, mainly because there aren't as many major cities in the Southern Hemisphere.

The thing about great circles is that the meridians are great circles, while the latitudes aren't (other than the Equator). This means that if we travel due north or south, we're following a great circle path, but not if we're traveling due east or west (unless we're on the Equator). It's traveling to airports far to the east or west that have great circle routes passing closer to the Poles.

Here's another way to intuit that the shortest distance between two points is a straight line. To begin, suppose I were to ask, what is the shortest path to the North Pole? The answer is obvious -- to get to the North Pole, we must travel

*north*. Going north takes us along a meridian -- a great circle path -- until we reach the Pole. So the shortest way to either Pole is obviously the great circle path.

But on a general sphere, there aren't two special points called the North and South Poles. Try taking an unmarked golf ball and declaring one of the points on its surface to be the "north pole." Now roll the ball a little, and then pick it up and try to locate the point that you declared to be the "north pole," and of course you'll be unable to find it.

Any point on the globe is a pole of many circles (using Legendre's definition of "pole"). There is nothing special about the actual Poles. So if the Poles have the property that the shortest path leading to them is the great circle path, then

*every*point must have that property, because

*any*point can be considered to be a pole. This is no different from coordinate proofs in Euclidean geometry -- we can place one vertex of a rectangle at the origin, since

*any*point on the plane can be considered the origin (and likewise we can place the axes along two sides of the rectangle).

If I were to get into a car and travel east without touching the steering wheel, I wouldn't be traveling east for very long. For without turning the wheel, the car will automatically travel a

*great circle path*, not the due east path. I would end up heading southeast approaching and passing the Equator, and keep going south until I reach the antipodes, at which point I start heading northeast until I returned to my starting point. Of course, there would be huge oceans in the way. An episode of

*Futurama*takes place on Mercury, where the main character Fry begins at the only gas station on the planet. So he starts driving in a great circle around the planet, but the car runs out of gas far from the only gas station, near the antipodes. A sign indicates that the gas station is 4,750 miles away no matter which way he drives.

We have one last proposition to cover -- Proposition 461:

461. The sum of the three sides of a spherical triangle is less than the circumference of a great circle.

Demonstration. Let

*ABC*be any spherical triangle, produce the sides

*AB*,

*AC*until they meet again in

*D*. The arcs

*ABD*,

*ACD*will be the semicircumferences of great circles, since two great circles always bisect each other; but in triangle

*BCD*the side

*BC*<

*BD*+

*CD*(459); adding to each

*AB*+

*AC*, we shall have

*AB*+

*AC*+

*BC*<

*ABD*+

*ACD*-- that is, less than the circumference of a great circle. QED

Notice that in this proof,

*A*and

*D*are antipodal points. Any two great circles bisect each other -- that is, they meet at antipodal points. This is combined with the Spherical Triangle Inequality to produce an upper bound on the perimeter of a triangle -- the perimeter of a spherical triangle must be less than that of a great circle.

The idea that any two great circles intersect in two points is important. We proved that the shortest distance between any two points on the globe is a great circle -- just as the shortest distance between any two points in the plane is a straight line. So great circles serve the same purpose in spherical geometry that lines do in Euclidean geometry -- as far as spherical geometry is concerned, great circles actually

*are*lines (and Legendre himself actually calls a great circle a "line" in his proof of Proposition 460 earlier).

So we can't have two parallel lines in spherical geometry unless we have two parallel great circles -- yet all great circles intersect. Therefore there are

*no*parallel lines in spherical geometry. In neutral geometry (Euclidean and hyperbolic), we can prove that parallel lines exist, but in spherical geometry, parallel lines don't exist. Therefore spherical geometry is not neutral (i.e., one in which the first four postulates of Euclid hold even if the fifth doesn't).

In neutral geometry, through any two points there is exactly one line (Euclid's first postulate). But in spherical geometry, through any two antipodal points there are infinitely many lines (since there are infinitely many great circles). Since the first postulate fails in spherical geometry, this geometry can't be neutral.

I've mentioned that Colorado and Wyoming aren't spherical polygons -- much less rectangles -- because two of their sides aren't great circle arcs. Here is an imaginary map in which both Colorado and Wyoming -- along with many other states -- really are spherical polygons (from the site that I linked to earlier when discussing the shapes of the states):

http://www.howderfamily.com/blog/closest-state-capital/

The states have been redrawn so as to minimize the distance to the state capital -- that is, the border between two states is equidistant from the state capitals. The locus of all points equidistant from two points is the perpendicular bisector plane (3D analog of the Perpendicular Bisector Theorem) -- which contains the center of the sphere (as the center is equidistant from all points). So the intersection of this plane and the sphere must be a great circle. Therefore if a state is landlocked (and doesn't border Canada or Mexico), the new shape has all great circle arcs as sides -- a true spherical polygon. (The author points out that there may be slight deviations from great circles, because the shape of the earth isn't a perfect sphere.)

Apparently my home is approximately halfway between Sacramento and Phoenix, since on this map I live right on the border between new California and new Arizona. (As a joke, the author mentions that this would all be easier if the earth were a cube -- recall one of the choices on the Calendar Reform list I posted six months ago was making the earth a cube!)

My next post will be in about a week.

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