Lesson 5-7 of the U of Chicago text discusses the sum of the angle measures in polygons, including triangles, quadrilaterals, and higher polygons. To me, this is the most arithmetic- and algebra-intensive lesson in all of the first semester.
The lesson begins with a discussion of Euclidean and non-Euclidean geometry. The 19th-century mathematician Karl Friedrich Gauss wanted to determine whether Euclidean geometry was true -- that is, that it accurately described the measure of the earth -- by experiment. The text shows a photo of three mountaintops that Gauss used as the vertices of a triangle, and the mathematician found that the sum of the angle measures of the triangle was, to within experimental error, 180 degrees.
Later on, the text states that if Gauss could have used a larger triangle -- say with one vertex at the North Pole and two vertices on the equator -- the angle-sum would have been greater than 180. The geometry of a sphere is not Euclidean, but is a special type of non-Euclidean geometry -- often called spherical geometry. As stated in the text:
"In a plane, two perpendiculars to the same line cannot intersect to form a triangle, but this can happen on a sphere. The surface of the earth can be approximated as a sphere. A triangle formed by two longitudes (north-south lines) an the equator is isosceles with two right base angles! Since there is a third angle at the North Pole, the measures add to more than 180 degrees. Thus neither the Two Perpendiculars Theorem nor the Triangle-Sum Theorem works on the surface of the earth."
But hold on a minute. It's obvious that the Triangle-Sum Theorem only holds in Euclidean geometry, as its proof uses the Alternate Interior Angles Consequence that depends on the Fifth Postulate. But we were able to prove the Two Perpendiculars Theorem on this blog, without using any sort of Parallel Postulate at all! So the Two Perpendiculars Theorem ought to hold for all types of geometry, both Euclidean and non-Euclidean -- yet it clearly doesn't hold for spherical geometry.
The truth is that spherical geometry differs from Euclidean geometry much more strongly than hyperbolic geometry differs from Euclidean. We can obtain hyperbolic geometry from Euclidean simply by dropping the Fifth Postulate and replacing it with an axiom stating that there are many parallels through a point not on the line. But we can't obtain spherical geometry in a similar way.
First of all, what exactly is a line in spherical geometry? (Recall that line is one of the undefined terms, so we can't rely on its definition.) Any figure that we think is a "line" on earth goes all the way around the world, and so is actually a circle. What we want is for a "line" to be the shortest distance between two points. Notice that smaller circles on the globe clearly look curved, but larger circles that go around the world look like straight lines to a traveler. Therefore the most "linear" circle is the largest possible circle -- one that shares a center with the earth. This is called a great circle -- and this is why the example in the text mentions two longitudes and the equator -- these are great circles. But the so-called "parallels of latitude" are not great circles and so are not "lines" (geodesics).
Now what postulates does this spherical geometry violate? Notice that there are no parallel lines on the sphere, because any two great circles intersect. (Once again, note that "parallels of latitude" are not great circles.) Any two longitudes meet at the poles, and so the Unique Line Assumption part of the Point-Line-Plane postulate is violated -- through the poles there are infinitely many lines rather than just one.
But any two great circles that intersect at the North Pole must intersect at the South Pole. And any two great circles that intersect at one point intersect at the point directly opposite that point -- often called the antipodes, or antipodal point. So one way to avoid this problem is to declare that two antipodal points are actually one point. The resulting geometry is called elliptic geometry.
Yet elliptic geometry still violates the postulates. Here I link to David Joyce's website for more discussion of elliptic geometry:
http://aleph0.clarku.edu/~djoyce/java/elements/bookI/propI16.html
Notice that this is a link to the first theorem of Euclid that fails in elliptic geometry. It is the Triangle Exterior Angle Inequality Theorem, or TEAI. Dr. Franklin Mason follows Euclid and uses the TEAI to derive the Parallel Tests. The Parallel Tests do not hold in elliptic geometry (of course not, since they prove lines parallel and there are no parallel lines).
In discussing which of Euclid's five postulates that fail in elliptic geometry, the link above writes:
Elliptic geometry satisfies some of the postulates of Euclidean geometry, but not all of them under all interpretations. Usually, I.Post.1, to draw a straight line from any point to any point, is interpreted to include the uniqueness of that line. But in elliptic geometry a completed “straight line” is topologically a circle so that any pair of points on it divide it into two arcs. Therefore, in elliptic geometry exactly two “straight lines” join any two given “points.”
Also, I.Post.2, to produce a finite straight line continuously in a straight line, is sometimes interpreted to include the condition that its ends don’t meet when extended. Under that interpretation, elliptic geometry fails Postulate 2.
Both of these are essentially part of our Point-Line-Plane Postulate. So this is the postulate that we'd have to rewrite if we want elliptic or spherical geometry. Our other postulates still hold -- we can still measure angles, we can still perform reflections, and we still have plane separation (of course, we'd call these halves "hemispheres").
Notice that ironically, our Fifth Postulate still holds in spherical geometry. Of course, it's vacuously true -- there are no parallel lines, so any statement of the form "if lines are parallel, then..." or something about a line intersecting two parallel lines, is vacuously true. The Parallel Consequences are also vacuously true in spherical geometry. Playfair also holds, provided that we write it the way that Dr. M writes it on his site:
"Through a point not on a given line, there’s at most one line parallel to the given line."
(emphasis mine)
"At most one" allows for the possibility of zero parallel lines. Technically, this is the form of Playfair that we proved earlier this week -- we only showed that at most one parallel line exists. The proof that at least one parallel line exists uses rotations and is not valid in spherical geometry.
Some teachers believe that we should briefly introduce high school students to non-Euclidean geometry -- and usually spherical geometry is suggested as it describes the earth. This is opposite what a college non-Euclidean geometry class would do -- in college, the emphasis is usually on hyperbolic geometry because its theorems are more similar to those of Euclidean geometry.
But it's often interesting to discuss with students how spherical geometry affects the earth. A classic brainteaser often goes as follows:
http://www.murderousmaths.co.uk/books/bearpuz.htm
- A bear hunter sets out from camp and walks one mile south.
- He sees a bear and is about to shoot it.
- The bear grabs his gun and eats it.
- The hunter runs away one mile east.
- He then walks one mile north and gets back to his camp and changes his underwear.
- What colour was the bear?
The answer is that the "colour" (sorry -- this is obviously from a British website) of the bear is white, since the puzzle describes a polar bear at the North Pole. Technically, this is not a spherical triangle, since the "one mile east" is along a parallel of latitude, not a great circle. It's not even close to being a great circle -- if the hunter ran approximately six miles east he would have walked in a complete circle around the pole.
Here's another puzzle related to spherical geometry. I've tutored students who've taken a long transoceanic flight, from California to Seoul, South Korea. Along the way, the plane ends up flying very close to Alaska. The question is, why does it fly so close to Alaska, rather than take a more sensible route closer to, say, Hawaii? The answer is that the flight near Alaska is actually shorter -- the flight follows a great circle, and the great circle through California and Korea passes near Alaska.
One final related question -- any two great circles meet at two antipodal points. Where exactly is the point on the globe that is antipodal to where we are standing now? Despite all the talk about "digging a hole to China," that country is not antipodal to the United States. As it turns out, most of the Lower 48 United States are not antipodal to land at all. If one dug a straight hole through the center of the earth starting anywhere in California, we'd end up in the Indian Ocean. But Hawaii is antipodal to parts of Botswana and Namibia in Africa, and of course Alaska is antipodal to Antarctica.
Here is a link to a map that calculates antipodes:
http://www.findlatitudeandlongitude.com/antipode-map/
Returning to Euclidean geometry, here's the proof of the Triangle-Sum Theorem given in the U of Chicago text. Since the book gives a two-column proof, I'll convert it to a paragraph proof:
Triangle-Sum Theorem:
The sum of the measures of the angles of a triangle is 180 degrees.
Given: Triangle ABC
Prove: angle A + angle B + angle C = 180
Proof:
Draw line BD with the measure of angle 1 (ABD) equal to angle A. By the Alternate Interior Angles Test, lines BD and AC are parallel. Then angle 3 (the angle on the other side of BC -- the text doesn't name it, but we can call it CBE if E is a point such that BE and BD are opposite rays) has the same measure as angle C, by the Alternate Interior Angles Consequence. By the Angle Addition Postulate, angles 1, 2 (ABC), and 3 add up to 180 degrees. Substituting, we get that angles A, ABC, and C add up to 180 degrees. QED
Right now, I am a substitute teacher, but last year I interviewed for a position as a regular teacher, and one of the things I was asked to prove was the Triangle-Sum Theorem. (I also had to derive the Quadratic Formula.) I gave a two-column proof similar to the one given in the text, and the principal told me that it was satisfactory, but that he might have preferred something like this:
Statements Reasons
1. Draw line BD parallel to line AC 1. Uniqueness of Parallels (Playfair)
2. angle 1 = angle A, angle 3 = angle C 2. Alternate Interior Angles Consequence
3. angle 2 = angle ABC 3. Reflexive Property of Equality
4. angle 1 + angle 2 + angle 3 = 180 4. Angle Addition Postulate
5. angle A + angle ABC + angle C = 180 5. Substitution (steps 2 and 3 into step 4)
So we include step 3, to show students that we are making three substitutions. Calling the same angle by two different names -- angle 2 and angle ABC -- emphasizes the need for a Reflexive Property to show that this angle equals itself. The U of Chicago just changes angle 2 to angle ABC without any explanation whatsoever. On the other hand, the U of Chicago distinguishes between the Angle Addition Postulate and the Linear Pair Theorem (which is the just the Angle Addition Postulate in the case that the angles add up to 180). The hope is that this form of the proof is the best for students to understand, which is our goal.
The Quadrilateral- and Polygon-Sum Theorems are just corollaries of the Triangle-Sum Theorem, as we expect. As I mentioned earlier, calculating (n - 2)180 (and dividing by n to find each angle of a regular polygon) is the most complicated algebra that I want students to have to do in first semester of the geometry class.
http://www.findlatitudeandlongitude.com/antipode-map/
Returning to Euclidean geometry, here's the proof of the Triangle-Sum Theorem given in the U of Chicago text. Since the book gives a two-column proof, I'll convert it to a paragraph proof:
Triangle-Sum Theorem:
The sum of the measures of the angles of a triangle is 180 degrees.
Given: Triangle ABC
Prove: angle A + angle B + angle C = 180
Proof:
Draw line BD with the measure of angle 1 (ABD) equal to angle A. By the Alternate Interior Angles Test, lines BD and AC are parallel. Then angle 3 (the angle on the other side of BC -- the text doesn't name it, but we can call it CBE if E is a point such that BE and BD are opposite rays) has the same measure as angle C, by the Alternate Interior Angles Consequence. By the Angle Addition Postulate, angles 1, 2 (ABC), and 3 add up to 180 degrees. Substituting, we get that angles A, ABC, and C add up to 180 degrees. QED
Right now, I am a substitute teacher, but last year I interviewed for a position as a regular teacher, and one of the things I was asked to prove was the Triangle-Sum Theorem. (I also had to derive the Quadratic Formula.) I gave a two-column proof similar to the one given in the text, and the principal told me that it was satisfactory, but that he might have preferred something like this:
Statements Reasons
1. Draw line BD parallel to line AC 1. Uniqueness of Parallels (Playfair)
2. angle 1 = angle A, angle 3 = angle C 2. Alternate Interior Angles Consequence
3. angle 2 = angle ABC 3. Reflexive Property of Equality
4. angle 1 + angle 2 + angle 3 = 180 4. Angle Addition Postulate
5. angle A + angle ABC + angle C = 180 5. Substitution (steps 2 and 3 into step 4)
So we include step 3, to show students that we are making three substitutions. Calling the same angle by two different names -- angle 2 and angle ABC -- emphasizes the need for a Reflexive Property to show that this angle equals itself. The U of Chicago just changes angle 2 to angle ABC without any explanation whatsoever. On the other hand, the U of Chicago distinguishes between the Angle Addition Postulate and the Linear Pair Theorem (which is the just the Angle Addition Postulate in the case that the angles add up to 180). The hope is that this form of the proof is the best for students to understand, which is our goal.
The Quadrilateral- and Polygon-Sum Theorems are just corollaries of the Triangle-Sum Theorem, as we expect. As I mentioned earlier, calculating (n - 2)180 (and dividing by n to find each angle of a regular polygon) is the most complicated algebra that I want students to have to do in first semester of the geometry class.
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