1. Introduction
2. A Single Postulate
3. What About Similarity?
4. What About the Third Dimension?
5. An Example in Neutral Geometry
6. Spherical AAS and Euclidean SSA
7. My Summer Plans for Spherical Geometry
8. Traditionalists: SteveH's June 15th Comment
9. Traditionalists: Barry Garelick's June 19th Post
10. Conclusion
Introduction
Today is the summer solstice -- the first official day of summer. I promised that I wouldn't write about Hofstadter's book in the summer, and now that the season is here, we're done with his book.
Instead, I wish to return to one of my favorite summer topics -- spherical geometry. Last summer, we finished reading Glen Van Brummelen's book on spherical trigonometry. Back in my July 12th post, I wrote how I wasn't sure whether I'd continue blogging about spherical geometry once we were done with Van Brummelen's book.
But as I mentioned last month, some of my most popular posts during the previous year were my Van Brummelen spherical trig posts. And I definitely enjoy the subject, so I do want to go back to writing about it now that school is out.
This is a blog about Common Core Geometry -- and of course, the only Geometry that appears in the Common Core is ordinary Euclidean geometry. So why do I want to mention spherical geometry on a Common Core blog, anyway?
Well, here is my vision for the ideal high school Geometry class. For most of the year, I would teach Euclidean geometry. But then at the end of the year -- after all state testing is completed, just for fun, I would introduce a little spherical geometry. I wrote that there exist some high schools that actually do this -- teach spherical geometry at the end of the year.
The idea is to compare Euclidean and spherical geometry. Now in theory, I wouldd want to prove as many theorems as possible that are valid in both Euclidean and spherical geometry. Then I'd want to introduce postulates that distinguish Euclidean from spherical geometry. Next, the rest of Euclidean geometry is taught using all of the postulates. Finally, when it's time for spherical geometry, I'd change the postulates and develop a few key spherical results.
There is a third type of geometry, called hyperbolic geometry. It turns out that the idea of proving as many results that hold in both Euclidean and a non-Euclidean geometry first before focusing only on Euclidean geometry is much easier when that Euclidean geometry is hyperbolic, not spherical. So why do I want to introduce spherical geometry?
Well, that's easy -- because the earth is approximately a sphere. Hyperbolic geometry doesn't appear to describe anything in the real world, even though some scientists suspect that the universe might actually be hyperbolic. On the other hand, spherical geometry has obvious applications to travel here on the round earth. This, indeed, was the subject of Van Brummelen's book.
And this is why spherical geometry at the end of the high school year is my goal. We can easily ask questions that motivate the use of spherical geometry, such as "Why do planes traveling between Asia and the lower 48 states go close to Alaska?" and even "What color is the bear?"
But I'd devoted many posts to this end over the years. Many of them are dead-ends -- at the time I thought that they were effective ways of teaching spherical geometry, but they weren't. Today's post is a summary of what I've achieved in spherical geometry thus far.
A Single Postulate
It's well-known that a single postulate distinguishes Euclidean from non-Euclidean geometry -- the Parallel Postulate. Lesson 13-6 of the U of Chicago text paraphrases Euclid's fifth postulate:
5. If two lines are cut by a transversal, and the interior angles on the same side of the transversal have a total measure of less than 180, then the lines will intersect on that side of the transversal.
The idea is that by replacing the fifth postulate with its negation, we get non-Euclidean geometry. The two types of geometry have the first four postulates of Euclid in common. The theorems that depend only on the first four postulates are called neutral geometry -- that is, they are neutral regarding the fifth postulate vs. its negation.
But unfortunately, this only works for hyperbolic geometry. It turns out that the first four postulates don't hold in spherical geometry, and thus spherical geometry isn't neutral. For example, we only need to look at Euclid's first postulate:
1. Two points determine a line segment.
This postulate is clearly false in spherical geometry. There isn't just a single line, or great circle, passing through the North and South Poles -- there are indeed infinitely many such lines.
In fact, as I often point out, Euclid's fifth postulate actually holds in spherical geometry. Again, we state Euclid's fifth postulate:
5. If two lines are cut by a transversal, and the interior angles on the same side of the transversal have a total measure of less than 180, then the lines will intersect on that side of the transversal.
And of course the lines will intersect on that side of the transversal, because all lines intersect. And indeed, they will intersect on both sides of the transversal, at a pair of antipodal points.
Of course, most Geometry texts don't state Euclid's fifth postulate. Instead, they usually quote a simpler parallel postulate, Playfair's postulate. But Playfair's postulate can be written as:
Through a point on a line, there is at most one line parallel to the given line.
Because of those words "at most," the postulate holds in spherical geometry. Once again, there are no parallel lines in spherical geometry.
Let's summarize what we've seen so far. To axiomatize hyperbolic geometry, we simply replace Playfair with its negation. To axiomatize neutral geometry, we simply omit both Playfair and its negation and include all other Euclidean axioms. But how do we axiomatize spherical geometry?
We know that we must negate Euclid's first postulate. In the U of Chicago text, Euclid's first postulate is included as part of the Point-Line-Plane Postulate, Lesson 1-7:
Point-Line-Plane Postulate:
(a) Unique line assumption: Through any two points, there is exactly one line.
Let's try replacing it with its negation:
(a') Through some two points, there isn't exactly one line.
But we can be more specific than this. First of all, through any two points on the sphere there's always at least one line -- the difference is that there could be more than one line:
(a') Through some two points, there is more than one line.
Moreover, through most pairs of points, there is exactly one line. It's only when the points are chosen to be antipodal can there be more than one line. So we might write:
(a') Through any two points, there is at least one line.
(a") Through any two antipodal points, there is more than one line.
This now requires us to define "antipodal points." I haven't completely figured out how this out yet, or whether it's possible to write a better version of this postulate without the word "antipodal."
Another part of Point-Line-Plane that clearly needs rewriting for spherical geometry is the Number Line Assumption:
(c) Number line assumption: Every line is a set of points that can be put into a one-to-one correspondence with the real numbers, with any point on it corresponding to 0 and any other point corresponding to 1.
Technically this holds in spherical geometry in the Cantorian sense -- the cardinality of the set of points on a great circle is the cardinality of the continuum (the real numbers), and thus there does exist a bijection (one-to-one correspondence) between R and the points on a circle. But the intent of (c) is to define a distance, and so we don't want an arbitrary bijection with R, but rather with some finite interval corresponding to the circumference of that great circle.
We tend to use the same units for circular distance as for angle measure (or arc measure), so we might try something like:
(c') Number line assumption: Every line is a set of points that can be put into a one-to-one correspondence with the interval [0, 360), with any point on it corresponding to 0.
We don't want any point to correspond to 1 -- we want a point that's one degree away from 0 to correspond to 1. By the way, it's usually standard to use radian measure instead:
(c') Number line assumption: Every line is a set of points that can be put into a one-to-one correspondence with the interval [0, 2pi), with any point on it corresponding to 0.
The reason for preferring radians (on the unit sphere) is that it makes area easier to calculate -- the area of a triangle is equal to its spherical excess -- the sum of its angle measures minus what that sum would have been in Euclidean geometry (that is, pi radians).
It might turn out to be better to assign multiple real numbers to a point instead of a single real -- for example, the point labeled 0 is also labeled 2pi, 4pi, 6pi, 8pi, -2pi, and so on. In any case, our choice can be reflected in the Distance assumption:
(d) Distance assumption: On a number line, there is a unique distance between points.
This might be changed so that there can be multiple distances between points -- the distance between the points labeled 0 and pi/3 isn't just pi/3, but also 7pi/3, 13pi/3, 19pi/3, and so on -- as well as 5pi/3 (the long way around the sphere), 11pi/3, 17pi/3, and so on.
I haven't completely decided how the new version of the Point-Line-Plane (Point-Circle-Sphere?) postulate will be written. Some decisions (antipodal points, degrees vs. radians) are still up in the air.
But the idea is that just as changing a single postulate (Playfair) is enough to make Euclidean into hyperbolic geometry, I want to change only Point-Line-Plane and obtain spherical geometry. All of the other postulates are to remain as they are, since I only want to change one postulate. This includes Playfair itself, since it actually holds in spherical geometry (due to "at most").
Let's look at the other axioms used in the U of Chicago text:
Triangle Inequality Postulate (This holds in spherical geometry. Actually on second thought, we'd need to double-check the distance given in parts c and d above to make sure that this still holds.)
Angle Measure Postulate (This holds in spherical geometry.)
Corresponding Angles Postulate (This fails in spherical geometry. But technically this postulate isn't actually needed for Euclidean geometry.)
Parallel Lines Postulate (This is replaced with Playfair, which holds in spherical geometry "at most.")
Reflection Postulate (This holds in spherical geometry.)
Area Postulate
This is a tricky one. One part of this postulate clearly seems to refer to Euclidean geometry:
(b) Rectangle Formula. The area of a rectangle with dimensions l and w is lw.
Notice that rectangles exist only in Euclidean geometry. Yet this part of the postulate is nonetheless valid in spherical geometry, because it is vacuously true. It asserts:
(b) If a quadrilateral is a rectangle with dimensions l and w, then its area is lw.
On the sphere, the antecedent of this conditional is always false. Thus the conditional itself is true, as it has no counterexamples. (There are zero rectangles on the sphere, and all of them have area lw. In other words, it's impossible to find a rectangle at all, much less one whose area isn't lw.)
The rest of the axiom is valid in spherical geometry. It asserts that spherical polygons have areas, though it no longer shows us how to find them.
What About Similarity?
We know that on the sphere, all similarity transformations are isometries, and so it's impossible for two figures to be similar without being congruent. Many texts list an AA (or other) Similarity Postulate, and we'd be forced to admit that this is a postulate that's spherically invalid.
But in our text, AA and the other similarity statements are all theorems. And since there isn't any similarity postulate, we're still true to our goal that Point-Line-Plane be the only postulate we change for the sphere.
Proving similarity without a similarity postulate is tricky. In past posts, I suggested a Dilation Postulate that's not unlike our Reflection Postulate. Such a postulate would be spherically invalid, and so we can't use it and maintain our goal that Point-Line-Plane be the only postulate we change.
We've seen several attempts to prove similarity without a postulate. For example, Hung-Hsi Wu proposes generalizing the Midpoint Connector Theorem. But his proof is too complex to use in a high school classroom.
Meanwhile, EngageNY, the Common Core curriculum used in New York, uses triangle area to prove the Side-Splitting Theorem. Then this theorem is used to prove the properties of dilations.
Our own U of Chicago text doesn't have a similarity postulate. Instead, in Lesson 12-1, there is a coordinate proof of the properties of the transformation S_k -- the transformation that maps the point (x, y) to the point (kx, ky). Clearly, this is a dilation centered at the origin with scale factor k. Then these properties are extended to all dilations -- in a coordinate proof, we can choose any point on the Euclidean plane to be the origin, so of course we choose the center of the dilation to be the origin.
This coordinate proof requires both the Distance and Slope Formulas. Notice that Lesson 8-7 has already given a proof of the Pythagorean Theorem using area, and Lesson 11-2 extends this theorem to the Distance Formula. But the Slope Formula is stated in Lesson 3-4 without proof. (The justification is basically, you already learned this in Algebra I.)
It's actually possible to prove a Slope Formula using area -- thus making it just like the Side-Splitter and Pythagorean Theorem proofs. It's mentioned at the Cut-the-Knot website on the page "A Lipogrammatic Proof of the Pythagorean Theorem":
https://www.cut-the-knot.org/pythagoras/LipogrammaticProof.shtml
(Michael notes that this is also a nice way to show that the concept of slope is well defined without referring to similar triangles.)
The only problem here is that Common Core expressly states that the similarity proofs of Pythagoras and Slope are preferred:
CCSS.MATH.CONTENT.8.EE.B.6
Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b.
CCSS.MATH.CONTENT.HSG.SRT.B.4
Prove theorems about triangles. Theorems include: a line parallel to one side of a triangle divides the other two proportionally, and conversely; the Pythagorean Theorem proved using triangle similarity.
On the other hand, notice that the Side-Splitting Theorem is also mentioned in the above standard, but it doesn't say "using triangle similarity." This is why the EngageNY text can first use area to prove Side-Splitter, then use Side-Splitter to derive the properties of similarity and dilations.
All of these proofs ultimately go back to either the area and existence of a rectangle, or the existence of parallel lines. This is why they are valid in Euclidean, but not spherical, geometry. As soon as we change Point-Line-Plane, the similarity results disappear.
What About the Third Dimension?
There is one postulate remaining in the U of Chicago text -- the Volume Postulate. But actually, we've said very little about the third dimension when it comes to spherical geometry.
Notice that even though a sphere is three-dimensional, spherical geometry is two-dimensional. Just as every point on the Euclidean plane requires only two coordinates to specify (x, y), every point on the sphere requires only two coordinates -- for example, latitude and longitude -- to specify.
But spherical geometry is said to represent the geometry of the earth -- and we who live on the earth are three-dimensional. It would appear that an obvious 3D analog of the spherical geometry would be to add altitude as the third dimension.
Now let's try to imagine what this would look like. So far, all of our lines turn out to be great circles on the sphere. But now we have new lines where the latitude and longitude remain constant and only the altitude changes. These lines actually look like Euclidean lines. So suddenly in our spherical geometry we have Euclidean lines.
But we also still have new lines that still look like great circles. These are great circles of spheres other than the unit sphere -- all of these spheres are concentric. Now consider two such circles -- the equator of the unit sphere, and the equator of one of the other spheres. These circles don't intersect -- indeed, considered as lines, we'd have to call them parallel. So we now have a "spherical" geometry where parallel lines exist!
So far, we've only considered lines that are either parallel or perpendicular to the unit sphere. But in Euclidean geometry, not only are there lines parallel or perpendicular to the xy-plane, there are also lines oblique to the xy-plane. So what would lines oblique to the unit sphere look like?
We consider the path of a rocket that is ascending while still orbiting at the equator. The path of such a rocket would be a spiral. Thus I suspect that our oblique lines would be spirals. There's more than one type of spiral, but I suspect that the equiangular or logarithmic spiral will serve our purpose. The equiangularity of these spirals makes it easier to find the angles between them -- since after all, we expect lines to intersect at angles.
In ordinary spherical geometry, we consider the North and South Poles to be special points, and the Equator to be a special line. Geographically these are special, but mathematically they aren't. If we wanted to, we could consider any other point to be the North Pole. (Indeed, there is another North Pole, called the North Magnetic Pole.) If I really wanted to, I could claim that the location of my house is in fact a North Pole. (Its antipodal point would be the South Pole.) Or instead, I could claim that a certain great circle passing through my house is in fact the Equator. This is similar, once again, to claiming that any given point on a Euclidean plane is in fact the origin.
But in our 3D model of "spherical" geometry, there is really is a point that is special -- the center of all the spheres. All the Euclidean lines seems to meet there, as do all the spirals.
Then again, logarithmic spirals are considered to get closer and closer to the origin without ever actually making it there. It might make more sense, in our model of "spherical" geometry, to define space as the set of all points other than the center, since this point is treated so much differently from any other point in space.
In some ways, this geometry acts like spherical geometry but in other ways it is Euclidean. It turns out that while there are three geometries in 2D (spherical, Euclidean, and hyperbolic), these generalize in different ways into 3D. This geometry we discovered is Euclidean-spherical, but there also are Euclidean-hyperbolic and spherical-hyperbolic geometries. A pure spherical geometry in 3D is actually modeled by the boundary of a 4D hypersphere. There we have no parallel lines, antipodal points, and everything we expect in a spherical geometry.
Since this is a quick introduction to spherical geometry at the end of the year in a Geometry class, it's just best to stick to 2D geometry. There's no reason to consider the Volume Postulate, or attempt to define volume, at all. When we consider spherical geometry, we are only look at the 2D examples.
An Example in Neutral Geometry
To achieve our goal, whenever we are given a statement to prove, we wish to give a proof that is valid in as many of the three geometries -- spherical, Euclidean, and hyperbolic -- as possible.
Let's start with the innocuous-looking statement:
The diagonals of a rectangle are congruent.
In which geometries can we prove this statement? Well, we see mention of a "rectangle," and we know that rectangles exist only in Euclidean geometry. And so we try a Euclidean proof.
Given: ABCD is a rectangle.
Prove: AC = BD.
Proof:
Statements Reasons
1. ABCD is a rectangle. 1. Given
2. Angles A, B right 2. Definition of rectangle
3. Angle A = B 3. All right angles are congruent.
4. ABCD is a pgram. 4. Quadrilateral Hierarchy Theorem
5. AD = BC 5. The pgram Consequences Theorem
6. AB = AB 6. Reflexive Property of Congruence
7. Triangle ABC = BAD 7. SAS Congruence [steps 5, 2, 6]
8. AC = BD 8. CPCTC
But hold on a minute. The statement which we are trying to prove:
The diagonals of a rectangle are congruent.
can be rewritten as:
If a quadrilateral is a rectangle, then its diagonals are congruent.
And we notice that this statement holds (albeit vacuously) in spherical and hyperbolic geometry. In other words, the statement to prove holds in all three geometries! So we should attempt to give a valid proof in all three geometries, or at least a second geometry besides Euclidean.
Let's start with neutral geometry, which combines Euclidean with hyperbolic geometry. Even though our ultimate goal is to study spherical geometry, we'll begin with neutral geometry because this concept has been studied more extensively than that of combining Euclidean with spherical geometry.
So far, the two-column proof above contains a step that is invalid in neutral geometry. This is Step 5, where we used the consequences of a pgram (opposite sides are congruent). In hyperbolic geometry, there exist parallelograms whose opposite sides are not congruent.
But then you might object, in Step 1, we are given that a rectangle exists. The existence of a rectangle already guarantees that we're in Euclidean geometry, so why are we wasting time with hyperbolic parallelograms whose opposite sides aren't congruent? So to complete our neutral proof, we add only an extra step between 1 and 2.
1.5 Euclidean geometry 1.5 If a rectangle exists, then the geometry is Euclidean.
This is now a neutrally valid proof, although it's more usual to write something like:
1.5 Playfair holds 1.5 If a rectangle exists, then Playfair holds.
5. AD = BC 5. If Playfair holds, then opposite sides of a pgram are congruent.
But this proof isn't elegant enough. Instead of a stating a neutral theorem that is vacuously true in one geometry and meaningfully true in the other, it's more elegant to state a theorem that is meaningfully true in both geometries. The idea is to generalize the idea of "rectangle" to something that actually exists in hyperbolic geometry, and then state and prove the theorem for both Euclidean rectangles and the hyperbolic analog of rectangles.
There are usually considered to be three generalizations of the rectangle in neutral geometry:
Given: ABCD is a Saccheri quad with Angles A, B right.
Area Postulate
This is a tricky one. One part of this postulate clearly seems to refer to Euclidean geometry:
(b) Rectangle Formula. The area of a rectangle with dimensions l and w is lw.
Notice that rectangles exist only in Euclidean geometry. Yet this part of the postulate is nonetheless valid in spherical geometry, because it is vacuously true. It asserts:
(b) If a quadrilateral is a rectangle with dimensions l and w, then its area is lw.
On the sphere, the antecedent of this conditional is always false. Thus the conditional itself is true, as it has no counterexamples. (There are zero rectangles on the sphere, and all of them have area lw. In other words, it's impossible to find a rectangle at all, much less one whose area isn't lw.)
The rest of the axiom is valid in spherical geometry. It asserts that spherical polygons have areas, though it no longer shows us how to find them.
What About Similarity?
We know that on the sphere, all similarity transformations are isometries, and so it's impossible for two figures to be similar without being congruent. Many texts list an AA (or other) Similarity Postulate, and we'd be forced to admit that this is a postulate that's spherically invalid.
But in our text, AA and the other similarity statements are all theorems. And since there isn't any similarity postulate, we're still true to our goal that Point-Line-Plane be the only postulate we change for the sphere.
Proving similarity without a similarity postulate is tricky. In past posts, I suggested a Dilation Postulate that's not unlike our Reflection Postulate. Such a postulate would be spherically invalid, and so we can't use it and maintain our goal that Point-Line-Plane be the only postulate we change.
We've seen several attempts to prove similarity without a postulate. For example, Hung-Hsi Wu proposes generalizing the Midpoint Connector Theorem. But his proof is too complex to use in a high school classroom.
Meanwhile, EngageNY, the Common Core curriculum used in New York, uses triangle area to prove the Side-Splitting Theorem. Then this theorem is used to prove the properties of dilations.
Our own U of Chicago text doesn't have a similarity postulate. Instead, in Lesson 12-1, there is a coordinate proof of the properties of the transformation S_k -- the transformation that maps the point (x, y) to the point (kx, ky). Clearly, this is a dilation centered at the origin with scale factor k. Then these properties are extended to all dilations -- in a coordinate proof, we can choose any point on the Euclidean plane to be the origin, so of course we choose the center of the dilation to be the origin.
This coordinate proof requires both the Distance and Slope Formulas. Notice that Lesson 8-7 has already given a proof of the Pythagorean Theorem using area, and Lesson 11-2 extends this theorem to the Distance Formula. But the Slope Formula is stated in Lesson 3-4 without proof. (The justification is basically, you already learned this in Algebra I.)
It's actually possible to prove a Slope Formula using area -- thus making it just like the Side-Splitter and Pythagorean Theorem proofs. It's mentioned at the Cut-the-Knot website on the page "A Lipogrammatic Proof of the Pythagorean Theorem":
https://www.cut-the-knot.org/pythagoras/LipogrammaticProof.shtml
(Michael notes that this is also a nice way to show that the concept of slope is well defined without referring to similar triangles.)
The only problem here is that Common Core expressly states that the similarity proofs of Pythagoras and Slope are preferred:
CCSS.MATH.CONTENT.8.EE.B.6
Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y = mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b.
CCSS.MATH.CONTENT.HSG.SRT.B.4
Prove theorems about triangles. Theorems include: a line parallel to one side of a triangle divides the other two proportionally, and conversely; the Pythagorean Theorem proved using triangle similarity.
On the other hand, notice that the Side-Splitting Theorem is also mentioned in the above standard, but it doesn't say "using triangle similarity." This is why the EngageNY text can first use area to prove Side-Splitter, then use Side-Splitter to derive the properties of similarity and dilations.
All of these proofs ultimately go back to either the area and existence of a rectangle, or the existence of parallel lines. This is why they are valid in Euclidean, but not spherical, geometry. As soon as we change Point-Line-Plane, the similarity results disappear.
What About the Third Dimension?
There is one postulate remaining in the U of Chicago text -- the Volume Postulate. But actually, we've said very little about the third dimension when it comes to spherical geometry.
Notice that even though a sphere is three-dimensional, spherical geometry is two-dimensional. Just as every point on the Euclidean plane requires only two coordinates to specify (x, y), every point on the sphere requires only two coordinates -- for example, latitude and longitude -- to specify.
But spherical geometry is said to represent the geometry of the earth -- and we who live on the earth are three-dimensional. It would appear that an obvious 3D analog of the spherical geometry would be to add altitude as the third dimension.
Now let's try to imagine what this would look like. So far, all of our lines turn out to be great circles on the sphere. But now we have new lines where the latitude and longitude remain constant and only the altitude changes. These lines actually look like Euclidean lines. So suddenly in our spherical geometry we have Euclidean lines.
But we also still have new lines that still look like great circles. These are great circles of spheres other than the unit sphere -- all of these spheres are concentric. Now consider two such circles -- the equator of the unit sphere, and the equator of one of the other spheres. These circles don't intersect -- indeed, considered as lines, we'd have to call them parallel. So we now have a "spherical" geometry where parallel lines exist!
So far, we've only considered lines that are either parallel or perpendicular to the unit sphere. But in Euclidean geometry, not only are there lines parallel or perpendicular to the xy-plane, there are also lines oblique to the xy-plane. So what would lines oblique to the unit sphere look like?
We consider the path of a rocket that is ascending while still orbiting at the equator. The path of such a rocket would be a spiral. Thus I suspect that our oblique lines would be spirals. There's more than one type of spiral, but I suspect that the equiangular or logarithmic spiral will serve our purpose. The equiangularity of these spirals makes it easier to find the angles between them -- since after all, we expect lines to intersect at angles.
In ordinary spherical geometry, we consider the North and South Poles to be special points, and the Equator to be a special line. Geographically these are special, but mathematically they aren't. If we wanted to, we could consider any other point to be the North Pole. (Indeed, there is another North Pole, called the North Magnetic Pole.) If I really wanted to, I could claim that the location of my house is in fact a North Pole. (Its antipodal point would be the South Pole.) Or instead, I could claim that a certain great circle passing through my house is in fact the Equator. This is similar, once again, to claiming that any given point on a Euclidean plane is in fact the origin.
But in our 3D model of "spherical" geometry, there is really is a point that is special -- the center of all the spheres. All the Euclidean lines seems to meet there, as do all the spirals.
Then again, logarithmic spirals are considered to get closer and closer to the origin without ever actually making it there. It might make more sense, in our model of "spherical" geometry, to define space as the set of all points other than the center, since this point is treated so much differently from any other point in space.
In some ways, this geometry acts like spherical geometry but in other ways it is Euclidean. It turns out that while there are three geometries in 2D (spherical, Euclidean, and hyperbolic), these generalize in different ways into 3D. This geometry we discovered is Euclidean-spherical, but there also are Euclidean-hyperbolic and spherical-hyperbolic geometries. A pure spherical geometry in 3D is actually modeled by the boundary of a 4D hypersphere. There we have no parallel lines, antipodal points, and everything we expect in a spherical geometry.
Since this is a quick introduction to spherical geometry at the end of the year in a Geometry class, it's just best to stick to 2D geometry. There's no reason to consider the Volume Postulate, or attempt to define volume, at all. When we consider spherical geometry, we are only look at the 2D examples.
An Example in Neutral Geometry
To achieve our goal, whenever we are given a statement to prove, we wish to give a proof that is valid in as many of the three geometries -- spherical, Euclidean, and hyperbolic -- as possible.
Let's start with the innocuous-looking statement:
The diagonals of a rectangle are congruent.
In which geometries can we prove this statement? Well, we see mention of a "rectangle," and we know that rectangles exist only in Euclidean geometry. And so we try a Euclidean proof.
Given: ABCD is a rectangle.
Prove: AC = BD.
Proof:
Statements Reasons
1. ABCD is a rectangle. 1. Given
2. Angles A, B right 2. Definition of rectangle
3. Angle A = B 3. All right angles are congruent.
4. ABCD is a pgram. 4. Quadrilateral Hierarchy Theorem
5. AD = BC 5. The pgram Consequences Theorem
6. AB = AB 6. Reflexive Property of Congruence
7. Triangle ABC = BAD 7. SAS Congruence [steps 5, 2, 6]
8. AC = BD 8. CPCTC
But hold on a minute. The statement which we are trying to prove:
The diagonals of a rectangle are congruent.
can be rewritten as:
If a quadrilateral is a rectangle, then its diagonals are congruent.
And we notice that this statement holds (albeit vacuously) in spherical and hyperbolic geometry. In other words, the statement to prove holds in all three geometries! So we should attempt to give a valid proof in all three geometries, or at least a second geometry besides Euclidean.
Let's start with neutral geometry, which combines Euclidean with hyperbolic geometry. Even though our ultimate goal is to study spherical geometry, we'll begin with neutral geometry because this concept has been studied more extensively than that of combining Euclidean with spherical geometry.
So far, the two-column proof above contains a step that is invalid in neutral geometry. This is Step 5, where we used the consequences of a pgram (opposite sides are congruent). In hyperbolic geometry, there exist parallelograms whose opposite sides are not congruent.
But then you might object, in Step 1, we are given that a rectangle exists. The existence of a rectangle already guarantees that we're in Euclidean geometry, so why are we wasting time with hyperbolic parallelograms whose opposite sides aren't congruent? So to complete our neutral proof, we add only an extra step between 1 and 2.
1.5 Euclidean geometry 1.5 If a rectangle exists, then the geometry is Euclidean.
This is now a neutrally valid proof, although it's more usual to write something like:
1.5 Playfair holds 1.5 If a rectangle exists, then Playfair holds.
5. AD = BC 5. If Playfair holds, then opposite sides of a pgram are congruent.
But this proof isn't elegant enough. Instead of a stating a neutral theorem that is vacuously true in one geometry and meaningfully true in the other, it's more elegant to state a theorem that is meaningfully true in both geometries. The idea is to generalize the idea of "rectangle" to something that actually exists in hyperbolic geometry, and then state and prove the theorem for both Euclidean rectangles and the hyperbolic analog of rectangles.
There are usually considered to be three generalizations of the rectangle in neutral geometry:
- Equiangular quadrilateral (all angles congruent)
- Saccheri quadrilateral (two adjacent right angles, congruent opposite sides with one adjacent to each right angle)
- Lambert quadrilateral (three right angles)
Of these three, the Lambert quadrilateral is most like the rectangle in that it has the most right angles, but the equiangular quadrilateral is most like the rectangle in terms of symmetry (and thus in terms of congruent parts). The Saccheri quadrilateral is intermediate on both counts.
It turns out that in neutral geometry both the equiangular and Saccheri quads have congruent diagonals, and so we can make our proof elegantly neutral by referring to either equiangular or Saccheri quads. The proof for Saccheri quads works out like clockwork:
Given: ABCD is a Saccheri quad with Angles A, B right.
Prove: AC = BD.
Proof:
Statements Reasons
1. ABCD is Saccheri. 1. Given
2. Angles A, B right 2. Definition of Saccheri quad
3. Angle A = B 3. All right angles are congruent.
4. ABCD is a pgram. 4. Quadrilateral Hierarchy Theorem
5. AD = BC 5. Definition of Saccheri quad
6. AB = AB 6. Reflexive Property of Congruence
7. Triangle ABC = BAD 7. SAS Congruence [steps 5, 2, 6]
8. AC = BD 8. CPCTC
In fact we can even drop Step 1 (we don't need to state that ABCD is Saccheri -- just state in the givens which angles are right and which sides are congruent) and Step 6 (a Saccheri quad is indeed a pgram, but we don't need it for the proof).
The proof that equiangular quads have congruent diagonals is more complex, since with only the congruence of angles known, we don't have the pair of congruent sides needed for SAS. A possible proof requires drawing in an auxiliary line -- the perpendicular bisector ofAB. We'll begin by naming the line m and the midpoint E, so that we have AE = BE. Let F be the point where m intersects CD.
But then suddenly, there's a problem. We don't know that m even intersectsCD. If we draw out a picture, it seems obvious that m should intersect CD, yet we can't take it for granted. I devoted several posts to the Euclidean proof that the perpendicular bisectors of a triangle are congruent -- the proof begins by assuming that two of them intersect and then showing that the third crosses there two. But the proof isn't neutral -- there are hyperbolic triangles whose perpendicular bisectors are parallel. So when it comes to neutral, we can never take line intersection for granted.
Here's my argument that m and line CD must intersect. We notice that since m is the perpendicular bisectorAB, it is the mirror of a reflection mapping A and B and AB to itself. Since angles A and B are congruent, this forces the mirror image of ray AD to be ray BC (though we don't yet know that the image of D is C).
Now in both Euclidean and hyperbolic geometry, a line can't enter a polygon without exiting it through one of the other sides. Since m enters the quad throughAB, it must exit the quad through one of the other sides, either AD, BC, or CD.
But sinceAD and BC are mirror images of each other, the mirror itself m can't intersect one of them without intersecting the other at that same point. In other words, the three lines AD, BC, and m are either parallel or concurrent. If they are parallel, then there's only one side left for m to intersect, which is the side we want, CD. If they are concurrent, then the point of concurrency must be on the opposite side of CD from AB -- otherwise ABCD wouldn't even be a polygon (as its sides would intersect at a point other than their vertex). Thus m must cross CD in order to reach that known point of concurrency. (A formal proof might name, for example, a Line Separation Postulate.)
This is rather messy -- and all of it is just to show that F, the intersection of m andCD, exists. Once we have that F exists, the proof suddenly becomes simple. I'll write it in paragraph format, since in two columns we'd have a missing reason:
Proof:
Statements Reasons
1. ABCD is equiangular. 1. Given
2. Let E midpointAB 2. Ruler Postulate
3. Let m perp. bis.AB 3. Protractor Postulate
4. m &CD intersect at F 4. ????
So let's pick up where we left off, where we find that the point F exists. We now consider the two Triangles AEF and BEF, and note that AE = BE (midpoint), Angles AEF = BEF (both right angles), and EF = EF (reflexive). Thus Triangles AEF and BEF are congruent by SAS, so AF = BF.
Now we consider Triangles ADF and BCF. We just found AF = BF, and we also have some congruent angles since ABCD is equiangular. We clearly have Angles D = C, and we also have DAF = CBF, found by subtracting EAF = EBF (CPCTC from earlier) from DAE = CBE (equiangular). Thus Triangles ADF and BCF are congruent by AAS, so AD = BC.
Once we have AD = BC, we can proceed just as in the Saccheri proof (which doesn't require the original base angles to be right angles, merely that they are congruent). Therefore we can finally conclude that AC = BD. QED
In fact, at this point we can make other conclusions about line m. We ultimately conclude that m is in fact the perpendicular bisector ofCD. It's amazing, yet frustrating, that once we prove that F exists, F and the original quad have all of these amazing properties. But I'm not quite sure that my reasoning that F exists is completely sound.
I did attempt a Google search for a full version of this proof. But most sources give only the proof for Saccheri quads (which is easy) and say very little about equiangular quads (except that some of them actually use the word "rectangle" to mean equiangular quad). If they do mention that equilateral quads have congruent opposite sides or diagonals, the proof is only left as an exercise.
I did find an alternate sequence of proofs that may avoid the problem with the existence of F. We begin with quad ABCD with Angles A = B, and AD = BC. Once again, it's easy to prove that the other two angles, C and D, are also congruent. (We can prove this once again by repeating the Saccheri proof, which only requires that the base angles are congruent.)
But then we ask, what happens if AD and BC are not congruent -- say AD < BC instead? Since BC is longer than AD, we can let G be the point between B and C such that AD = BG. Then by the previous reasoning, ABGD has congruent summit angles ADG and BGD. But Angle C doesn't equal BGD -- in fact, Angle C < BGD. This is because BGD is the exterior angle of Triangle CDG -- a result which we often referred to as the Triangle Exterior Angle Inequality (TEAI), which is valid in neutral geometry.
So if AD < BC then Angle C > D -- and likewise if AD > BC then Angle C < D. We have proved that if the base angles of a quad are congruent yet the opposite sides adjacent to these angles aren't, then the summit angles aren't congruent -- the smaller summit angle is farther from the shorter side.
And so it follows that if we're given a quad with both the base and summit pairs congruent (as they are in an equiangular quad with all four angles congruent), then the opposite sides (one side between each base and summit angle) must be congruent -- otherwise the summit angle farther from the shorter side would be smaller. It's similar to the derivation of Unequal Angles from Unequal Sides in Lesson 13-7 (and indeed, both proofs begin with TEAI).
So once again, we conclude that AD = BC, from which we proceed as in the Saccheri case.
Let's think about why we are proving this theorem in the first place. The original theorem to prove, if you'll recall, is:
The diagonals of a rectangle are congruent.
I wrote that it would be elegant to prove this in neutral geometry by generalizing to quads that actually exist in hyperbolic geometry:
The diagonals of a Saccheri quad are congruent.
The diagonals of an equiangular quad are congruent.
Now the first proof was much easier than the second. So then, if all we wanted to do was generalize "rectangle" to something that exists in hyperbolic geometry, why even bother with equiangular quads and the more difficult proof? We can just generalize "rectangle" to "Saccheri quad" and be done.
Well, it's actually not quite obvious that a rectangle even is a Saccheri quad. A rectangle is defined as a quad with four right angles. All right angles are congruent, and so it follows directly from the definition of a rectangle that it is equiangular. Likewise, it's evident that a rectangle is a Lambert quad, since a quad with four right angles obviously has (at least) three right angles.
It's only the statement that a rectangle is a Saccheri quad that doesn't follow trivially from the definition of rectangle, since that definition mentions nothing about congruent sides. Of course, it's exactly that given pair of congruent sides that makes the Saccheri quad proofs easy.
Thus in order to use the Saccheri quad generalization, we must prove that a rectangle is in fact a Saccheri quad. And there are two ways to do this -- either go back to "since a rectangle exists, we must be in Euclidean geometry" and derive the congruent sides in Euclid, or repeat the same neutral proof that an equilateral quad (with a rectangle clearly is) has a pair of congruent sides.
Spherical AAS and Euclidean SSA
All of this refers to neutral geometry -- that is, Euclidean and hyperbolic geometry. So far, I haven't said anything about whether this proof is valid in spherical geometry.
My idea is to come up with a "natural geometry." This is Euclidean combined with spherical geometry, in the same way that "neutral geometry" combines Euclidean and hyperbolic geometry.
So far, I said that I want all postulates of planar Euclidean geometry to remain (including Playfair) except for Point-Line-Plane. That is, all planar Euclidean postulates would be part of my vision for this "natural geometry." It will probably be helpful to have a "natural Point-Line-Plane Postulate" that assumes those facts about points and lines that hold in both Euclidean and spherical geometries. It might start like this:
Natural Point-Line-Plane Postulate:
(a) Through any two points, there is at least one line.
(b) Through any three points, there is at most one line.
etc.
and it would state other basic facts about both geometries, such as the existence of distance.
Now let's return to the statement that we've been trying to prove:
The diagonals of a rectangle are congruent.
This statement holds in both spherical (albeit vacuously) and Euclidean geometry, so we should be able to find a proof in our new natural geometry. And once again, we don't want to start with "since a rectangle exists, we must be in Euclidean geometry" -- instead, we want to prove that some generalization of rectangles exist and have congruent diagonals in both geometries.
It turns out that equiangular, Saccheri, and Lambert quadrilaterals all exist on the sphere. And the previously given proof that Saccheri quads have congruent diagonals is spherically valid. Thus we already have a natural proof of "the diagonals of a Saccheri quad are congruent." But once again, it's not obvious in natural geometry that a rectangle (which only exists in Euclid) is in fact Saccheri. So once again, it's preferable to find a proof of "the diagonals of an equiangular quad are congruent."
Returning to my first proof above, the problem with the existence of F vanishes on the sphere -- all lines intersect on the sphere, so of course m and line CD must intersect! But that proof is invalid -- it uses AAS, which fails on the sphere. And our alternate proof which avoids AAS is also invalid -- it uses TEAI, which also fails on the sphere.
I attempted a Google search for this proof, but of course I couldn't find it -- there's precious little written even on hyperbolic equiangular quads, let alone spherical ones. Since I don't have a proof of this statement, I also wonder whether it's in fact false -- there exist equiangular quads on the sphere with unequal opposite sides and unequal diagonals. But I couldn't find such a counterexample either.
The lack of AAS in spherical geometry reminds me of the lack of SSA in Euclidean geometry. The U of Chicago text mentions that there's a special case of SSA which does hold in Euclidean geometry -- called SsA, it refers to the case where the known congruent angles are opposite the longer of the known congruent sides. My Second Edition doesn't give a proof of SsA, but I notice that a complex proof does appear in the modern Third Edition of the text.
We might notice that in the triangles we're hoping to prove congruent by AAS (ADF and BCF), we notice that the known congruent sides at AF = BF. The congruent angles opposite these sides, D and C, are known to be larger than the other two angles DAF and CBF. (Once again, we know this because DAF is a subangle of DAE, so DAF < DAE = C by the equiangularity of quad ABCD.) Thus even without AAS, if there could be a spherical AaS (much like Euclidean SsA), then the above proof becomes valid and we will have indeed shown the symmetry of equiangular quads on the sphere.
I don't recall the full proof of SsA in the third edition of the text. But SSA does remind me of the ambiguous case of the Law of Sines. We have:
a/sin A = b/sin B
Now suppose a, b, and A are all given, so a is opposite the known angle A. We notice that if a > b (the SsA case), then there is only one possible triangle. To see why, we solve for sin B:
sin B = (b sin A)/a
Now there are two possible angles with the same sine -- one acute and one obtuse. Now we just appeal to Unequal Sides -- if a > b, then A > B. If B were obtuse, then so would A, since A > B. So Triangle ABC would have two obtuse angles, a contradiction (in Euclidean geometry). Thus there is only one possible angle for B, the acute angle. We can now completely solve the triangle. Since a triangle is completely determined by SsA, this implies the SsA Congruence Theorem. QED
I don't remember the proof of SsA from the Third Edition of the text, but I know that it doesn't simply appeal to the Law of Sines, which isn't taught in any edition of the Geometry text. But it might give us a hint as to whether AaS is valid on the sphere or not.
We recall from Van Brummelen the past two summers that there is a spherical Law of Sines. In fact there are many trig formulas in that text. Most of them I have trouble remembering, but the spherical Law of Sines is one of the easiest to recall. It's just like the Euclidean Law of Sines, except that everything in the spherical formula requires the sine:
sin a/sin A = sin b/sin B
Now suppose A, B, and a are all given. We now solve for sin b:
sin b = (sin B sin a)/sin A
And while there are statements similar to Unequal Sides for spherical triangles, we realize that triangles can indeed have two "obtuse" sides (that is, longer than a quadrant). Thus knowing that both A > B and a > b doesn't allow us to rule out that b > 90. Thus AaS remains ambiguous.
Unfortunately, Van Brummelen has very little to say about AAS in his book. He writes more about both ASA and even SSA than AAS. I'm actually now starting to suspect that rather than AaS, it might actually be aAS (where the angle opposite the known side is the smaller angle) that can be fully determined on the sphere. If this is true, then my proof that requires AaS is surely invalid. Once again, I don't know whether there is a counterexample to our equiangular quad claim or not.
My Summer Plans for Spherical Geometry
Let me reveal what exactly my plans are for spherical geometry this summer. Note that not every post this summer will be about spherical geometry -- here I'm only referring to the spherical posts.
Each spherical post will be labeled as a "unit" -- as in a hypothetical unit of a high school course that begins with Euclidean geometry and ends with an intro to spherical geometry. This means that most of the posts will actually be about Euclidean geometry from another perspective.
The first such post will introduce our most important postulate -- Point-Line-Plane. In that post, I want to give the Euclidean (same as the U of Chicago), spherical, and natural versions. Once again, the "natural" Point-Line-Plane postulate is supposed to capture statements about points and lines that hold in both geometries.
In subsequent posts, we'll try to prove as many statements as possible using natural Point-Line-Plane as well as the other postulates of Euclidean geometry (including Playfair). We'll keep on emphasizing transformations just as in Common Core, including reflections and rotations. Translations might not necessary exist in spherical geometry.
Notice that in an actual class, we'd state only Euclidean Point-Line-Plane. I'd be careful to use only the parts of Point-Line-Plane that are natural, but we don't tell the students this. Only at the end when we introduce spherical geometry do we reveal which statements hold in both geometries.
At this point you might wonder, what the point of making these posts? For one thing, we know that spherical geometry was first created by Bernhard Riemann, yet very little writing from past two centuries since the Riemann sphere can be found online. My hope is that I can post and provide many of the proofs that I can't locate online.
Traditionalists: SteveH's June 15th Comment
I've added the "traditionalists" label to this post, since the traditionalists have been active lately.
In my last post, I linked to our favorite traditionalist, Barry Garelick, and his Flag Day post along with several comments. At the time, our favorite commenter SteveH hadn't posted yet. So let's discuss SteveH's comment right now:
https://traditionalmath.wordpress.com/2019/06/14/misunderstandings-about-understanding-dept/
SteveH:
Answer getting also requires a lot of understanding at the lowest levels of math, including the times table. What is 6*7? 42, isn’t it? Well 5*7=35 + 7 = 42, because 7 is 5+2 and I like to add by fives. Nobody does math only with rote memory. Then there are things like telling time and counting change with quarters, dimes, nickels, and pennies.
This is fine and dandy. Indeed, I like the idea of teaching 6 * 7 = 5 * 7 + 7 = 35 + 7 = 42. Indeed, we discussed this earlier on the blog regarding learning times tables in other bases. For example, since 6 is half of the dozenal base 12, multiplying by dozenal 7 is a bit like multiplying by decimal 6, and SteveH's trick here. A dozenalist would thus write 6 * 7 = 6 * 6 + 6 = 30 + 6 = 36 (dozenal).
But here's the problem. A third grade teacher teaches 6 * 7 = 5 * 7 + 7 = 35 + 7 = 42. A student has already memorized 6 * 7 = 42, and thus omits the "5 * 7 + 7 = 35 + 7 =" part. The teacher marks the student wrong, or deducts points, for omitting the "5 * 7 + 7 = 35 + 7 =" part. And the first to complain would be SteveH and the other traditionalists.
SteveH:
Most people forget the incremental understanding process they went through to get there. Telling time is very difficult, what with 365 days per year (plus leap year), 30 or so days per month (do we stop teaching kids the rote rhyme? I don’t even know the reason or understanding of the rhyme.), 24 hours per day (AM and PM), 60 minutes per hour, and 60 seconds per minute. Do they have to know about the solar system, years for different planets, spin on a tilted axis to the sun, and why 24 and 60 as bases? Do they have to know about different bases? My son had a thematic unit in first grade about little Eskimo kids in Alaska without ever knowing why it’s so cold and dark up there most of the year. Clearly, they aren’t worried about rote knowledge.
This is a tricky one. My priority is to address and avoid "Why?" questions asked by actual students sitting in actual classrooms. SteveH's first grade son probably never asked "Why is it so cold and dark up in Alaska most of the year?" -- thus the first grade teacher never had to explain it.
Actual students won't ask why there are 60 minutes per hour and 60 seconds per hour. Instead, they will ask "Why do we have to learn to tell time (on an analog clock)?" Many of them believe that the existence of digital time on cell phones renders learning analog time archaic. And some of them might even think that counting change with quarters, dimes, etc., is also archaic, because inflation has resulting in items costing more than a quarter these days -- also because they plan to do nearly 100% of all transactions with a credit or debit card anyway.
Some students might ask why 6 * 7 is 42. But others might ask why they need to learn their sixes and sevens times tables, since they believe that the existence of calculators renders memorizing the times tables archaic. What we wish to do is reach students who refuse to memorize that they believe that they shouldn't have to.
SteveH:
“Answer getting” is something they made up just to push their beliefs. So what’s driving it? All of that silliness disappeared (for the most part) when my son got to high school. It’s a K-8 thing and in our schools, the fundamental driving force in K-8 is full academic inclusion. Our high school, however, is full environmental inclusion. To get full academic inclusion to work, however, requires lowering grade level expectations and creating a world of magic natural differentiated learning where “kids will learn when they are ready.”
I've seen that phrase "full inclusion" mentioned by several traditionalists, not just SteveH. All I can glean from this posts is that "full inclusion" is bad. I'd think that "full inclusion" would be good because its negation, "someone is excluded" is bad. But to traditionalists, "full inclusion" is bad.
Here SteveH also distinguishes between "full academic inclusion" and "full environmental inclusion," with the former being definitely bad. I don't yet know (or understand) what the difference is. He mentions "full inclusion" once more in this post:
SteveH:
Traditionalists: Barry Garelick's June 19th Post
On Wednesday, Garelick himself posted again:
https://traditionalmath.wordpress.com/2019/06/19/my-first-educational-treatise/
My mother was my first critic of my first educational treatise, written when I was a senior in high school. During my junior year, I had tried unsuccessfully to get transferred out of a geometry class taught by a teacher whose reputation had preceded her for years.
Since this is a Geometry blog, I definitely want to read what Garelick has to say about Geometry.
In retrospect, she was a very bright woman who had emotional problems and spent the class talking about this, that and anything other than about geometry and left it to us to read the book, do the problems, and make presentations on the board. At that time, however, I wanted someone who taught.
Much of the debate regarding traditionalism vs. discovery math doesn't involve Geometry, but we definitely know of one Geometry text that's discovery -- Michael Serra's Discovering Geometry. I don't know whether Garelick's Geometry teacher used Serra's text, but we gather that based on the way she taught her class, it might as well be Discovering Geometry. She had her students discover geometric relations for themselves as opposed to teaching them the postulates and theorems.
In the end, Garelick wrote a pamphlet about how bad his teacher was, and even sold it:
I started selling the pamphlets at school for 25 cents a piece which got me in a little bit of trouble at school, but that’s another story. One day my mother got mad at me for something I said to her (probably complaining about something I didn’t like about something she did) and she let me have it. It was one of those “After all I’ve done for you” type rants, and once she got going there was no stopping her. Among the things she mentioned was running off the dittos and helping me glue them together.
See -- that's something you could have bought for a quarter back in the day. I wonder how much he would have sold it for if he were a young high school student nowadays.
Anyway, Garelick went on in the pamphlet to criticize English teachers and their grading policies. As this is a math blog, I don't comment on grades in English classes here.
He now moves forward to his teaching career, and tells a story about his previous middle school:
In that particular school, I was available for tutoring students prior to the first period, and some of her students would occasionally come in for help. In one case, a boy was having difficulty with proportion problems, such as 3/x = 2/5. I showed him how it is solved using cross multiplication (without explaining why cross multiplication works—I was after just getting him through the assignment). I ran into the boy’s father during the summer, and he made a point to thank me for helping his son. I found this odd since I had only tutored him maybe one other time. “He was getting D’s and F’s on his test and since you helped him, he started getting A’s and he ended up with an A- in the class.”
Proportions often come up in the traditionalists' posts. Since this story takes place in Garelick's past, it's possible that this was before the advent of Common Core. Nevertheless, it's likely that his colleague used Common Core methods of teaching proportions in middle school -- including double number lines and tape diagrams -- and these were confusing the student in the story.
Let's compare how Common Core teaches proportions in middle school to how it teaches another middle school standard, the Pythagorean Theorem. The Common Core developers figured that many students consider equations to be huge turnoff, and thus the idea was to show students how to solve as many problems as possible without needed to write an equation. This is why they use double number lines and tape diagrams to solve proportions. On the other hand, even the Common Core writers know that there is no easier way to solve a Pythagorean Theorem problem other than to use the underlying equation, a^2 + b^2 = c^2. That's why there is no special "Common Core method" of solving a right triangle problem.
Garelick tells us that this boy he tutored found it much easier to solve 3/x = 2/5 by cross multiplying than any method that the boy's own teacher showed him. But how many other students sitting in that boy's class are confused by cross multiplication? How many of them might try to write 5x = 6 or even 3x = 10 instead of the correct 2x = 15? The purpose of the double number lines and tape diagrams is so the student can visualize why x must clearly be greater than 5, so that 5x = 6 (leading to x = 1.2) can't possibly be correct?
Conclusion
Garelick clearly wants solving proportions by cross multiplying to be taught earlier than it currently is under the Common Core. He doesn't tell us how old the boy in his story was -- as far as we know, he was an Algebra I eighth grader. Let's give the other teacher the benefit of the doubt that he was a seventh grader.
I was actually thinking about when proportion equations should be taught while researching spherical vs. Euclidean geometry and the similarity proofs. We see how the U of Chicago Geometry text states theorems about slope without proof (since slope was taught in Algebra I) and then uses slope in coordinate proofs about S_k.
Perhaps the best plan to meet the Common Core Standards is to teach the derivation of slope from similarity in Algebra I. (The problem in eighth grade is that in many classes, all of the EE standards, including slope, are taught before any G standards, including similarity.) Then since similarity is being discussed in Algebra I anyway, that's when solving proportion equations by cross multiplying should definitely occur -- and be emphasized more than they currently are in Algebra I. Of course, that won't be good enough for Garelick and the traditionalists.
I look forward to writing our Point-Line-Plane postulates for spherical and natural geometry. But that probably won't be in my next post, since they will take some time for me to devise.
Proof:
Statements Reasons
1. ABCD is Saccheri. 1. Given
2. Angles A, B right 2. Definition of Saccheri quad
3. Angle A = B 3. All right angles are congruent.
4. ABCD is a pgram. 4. Quadrilateral Hierarchy Theorem
5. AD = BC 5. Definition of Saccheri quad
6. AB = AB 6. Reflexive Property of Congruence
7. Triangle ABC = BAD 7. SAS Congruence [steps 5, 2, 6]
8. AC = BD 8. CPCTC
In fact we can even drop Step 1 (we don't need to state that ABCD is Saccheri -- just state in the givens which angles are right and which sides are congruent) and Step 6 (a Saccheri quad is indeed a pgram, but we don't need it for the proof).
The proof that equiangular quads have congruent diagonals is more complex, since with only the congruence of angles known, we don't have the pair of congruent sides needed for SAS. A possible proof requires drawing in an auxiliary line -- the perpendicular bisector of
But then suddenly, there's a problem. We don't know that m even intersects
Here's my argument that m and line CD must intersect. We notice that since m is the perpendicular bisector
Now in both Euclidean and hyperbolic geometry, a line can't enter a polygon without exiting it through one of the other sides. Since m enters the quad through
But since
This is rather messy -- and all of it is just to show that F, the intersection of m and
Proof:
Statements Reasons
1. ABCD is equiangular. 1. Given
2. Let E midpoint
3. Let m perp. bis.
4. m &
So let's pick up where we left off, where we find that the point F exists. We now consider the two Triangles AEF and BEF, and note that AE = BE (midpoint), Angles AEF = BEF (both right angles), and EF = EF (reflexive). Thus Triangles AEF and BEF are congruent by SAS, so AF = BF.
Now we consider Triangles ADF and BCF. We just found AF = BF, and we also have some congruent angles since ABCD is equiangular. We clearly have Angles D = C, and we also have DAF = CBF, found by subtracting EAF = EBF (CPCTC from earlier) from DAE = CBE (equiangular). Thus Triangles ADF and BCF are congruent by AAS, so AD = BC.
Once we have AD = BC, we can proceed just as in the Saccheri proof (which doesn't require the original base angles to be right angles, merely that they are congruent). Therefore we can finally conclude that AC = BD. QED
In fact, at this point we can make other conclusions about line m. We ultimately conclude that m is in fact the perpendicular bisector of
I did attempt a Google search for a full version of this proof. But most sources give only the proof for Saccheri quads (which is easy) and say very little about equiangular quads (except that some of them actually use the word "rectangle" to mean equiangular quad). If they do mention that equilateral quads have congruent opposite sides or diagonals, the proof is only left as an exercise.
I did find an alternate sequence of proofs that may avoid the problem with the existence of F. We begin with quad ABCD with Angles A = B, and AD = BC. Once again, it's easy to prove that the other two angles, C and D, are also congruent. (We can prove this once again by repeating the Saccheri proof, which only requires that the base angles are congruent.)
But then we ask, what happens if AD and BC are not congruent -- say AD < BC instead? Since BC is longer than AD, we can let G be the point between B and C such that AD = BG. Then by the previous reasoning, ABGD has congruent summit angles ADG and BGD. But Angle C doesn't equal BGD -- in fact, Angle C < BGD. This is because BGD is the exterior angle of Triangle CDG -- a result which we often referred to as the Triangle Exterior Angle Inequality (TEAI), which is valid in neutral geometry.
So if AD < BC then Angle C > D -- and likewise if AD > BC then Angle C < D. We have proved that if the base angles of a quad are congruent yet the opposite sides adjacent to these angles aren't, then the summit angles aren't congruent -- the smaller summit angle is farther from the shorter side.
And so it follows that if we're given a quad with both the base and summit pairs congruent (as they are in an equiangular quad with all four angles congruent), then the opposite sides (one side between each base and summit angle) must be congruent -- otherwise the summit angle farther from the shorter side would be smaller. It's similar to the derivation of Unequal Angles from Unequal Sides in Lesson 13-7 (and indeed, both proofs begin with TEAI).
So once again, we conclude that AD = BC, from which we proceed as in the Saccheri case.
Let's think about why we are proving this theorem in the first place. The original theorem to prove, if you'll recall, is:
The diagonals of a rectangle are congruent.
I wrote that it would be elegant to prove this in neutral geometry by generalizing to quads that actually exist in hyperbolic geometry:
The diagonals of a Saccheri quad are congruent.
The diagonals of an equiangular quad are congruent.
Now the first proof was much easier than the second. So then, if all we wanted to do was generalize "rectangle" to something that exists in hyperbolic geometry, why even bother with equiangular quads and the more difficult proof? We can just generalize "rectangle" to "Saccheri quad" and be done.
Well, it's actually not quite obvious that a rectangle even is a Saccheri quad. A rectangle is defined as a quad with four right angles. All right angles are congruent, and so it follows directly from the definition of a rectangle that it is equiangular. Likewise, it's evident that a rectangle is a Lambert quad, since a quad with four right angles obviously has (at least) three right angles.
It's only the statement that a rectangle is a Saccheri quad that doesn't follow trivially from the definition of rectangle, since that definition mentions nothing about congruent sides. Of course, it's exactly that given pair of congruent sides that makes the Saccheri quad proofs easy.
Thus in order to use the Saccheri quad generalization, we must prove that a rectangle is in fact a Saccheri quad. And there are two ways to do this -- either go back to "since a rectangle exists, we must be in Euclidean geometry" and derive the congruent sides in Euclid, or repeat the same neutral proof that an equilateral quad (with a rectangle clearly is) has a pair of congruent sides.
Spherical AAS and Euclidean SSA
All of this refers to neutral geometry -- that is, Euclidean and hyperbolic geometry. So far, I haven't said anything about whether this proof is valid in spherical geometry.
My idea is to come up with a "natural geometry." This is Euclidean combined with spherical geometry, in the same way that "neutral geometry" combines Euclidean and hyperbolic geometry.
So far, I said that I want all postulates of planar Euclidean geometry to remain (including Playfair) except for Point-Line-Plane. That is, all planar Euclidean postulates would be part of my vision for this "natural geometry." It will probably be helpful to have a "natural Point-Line-Plane Postulate" that assumes those facts about points and lines that hold in both Euclidean and spherical geometries. It might start like this:
Natural Point-Line-Plane Postulate:
(a) Through any two points, there is at least one line.
(b) Through any three points, there is at most one line.
etc.
and it would state other basic facts about both geometries, such as the existence of distance.
Now let's return to the statement that we've been trying to prove:
The diagonals of a rectangle are congruent.
This statement holds in both spherical (albeit vacuously) and Euclidean geometry, so we should be able to find a proof in our new natural geometry. And once again, we don't want to start with "since a rectangle exists, we must be in Euclidean geometry" -- instead, we want to prove that some generalization of rectangles exist and have congruent diagonals in both geometries.
It turns out that equiangular, Saccheri, and Lambert quadrilaterals all exist on the sphere. And the previously given proof that Saccheri quads have congruent diagonals is spherically valid. Thus we already have a natural proof of "the diagonals of a Saccheri quad are congruent." But once again, it's not obvious in natural geometry that a rectangle (which only exists in Euclid) is in fact Saccheri. So once again, it's preferable to find a proof of "the diagonals of an equiangular quad are congruent."
Returning to my first proof above, the problem with the existence of F vanishes on the sphere -- all lines intersect on the sphere, so of course m and line CD must intersect! But that proof is invalid -- it uses AAS, which fails on the sphere. And our alternate proof which avoids AAS is also invalid -- it uses TEAI, which also fails on the sphere.
I attempted a Google search for this proof, but of course I couldn't find it -- there's precious little written even on hyperbolic equiangular quads, let alone spherical ones. Since I don't have a proof of this statement, I also wonder whether it's in fact false -- there exist equiangular quads on the sphere with unequal opposite sides and unequal diagonals. But I couldn't find such a counterexample either.
The lack of AAS in spherical geometry reminds me of the lack of SSA in Euclidean geometry. The U of Chicago text mentions that there's a special case of SSA which does hold in Euclidean geometry -- called SsA, it refers to the case where the known congruent angles are opposite the longer of the known congruent sides. My Second Edition doesn't give a proof of SsA, but I notice that a complex proof does appear in the modern Third Edition of the text.
We might notice that in the triangles we're hoping to prove congruent by AAS (ADF and BCF), we notice that the known congruent sides at AF = BF. The congruent angles opposite these sides, D and C, are known to be larger than the other two angles DAF and CBF. (Once again, we know this because DAF is a subangle of DAE, so DAF < DAE = C by the equiangularity of quad ABCD.) Thus even without AAS, if there could be a spherical AaS (much like Euclidean SsA), then the above proof becomes valid and we will have indeed shown the symmetry of equiangular quads on the sphere.
I don't recall the full proof of SsA in the third edition of the text. But SSA does remind me of the ambiguous case of the Law of Sines. We have:
a/sin A = b/sin B
Now suppose a, b, and A are all given, so a is opposite the known angle A. We notice that if a > b (the SsA case), then there is only one possible triangle. To see why, we solve for sin B:
sin B = (b sin A)/a
Now there are two possible angles with the same sine -- one acute and one obtuse. Now we just appeal to Unequal Sides -- if a > b, then A > B. If B were obtuse, then so would A, since A > B. So Triangle ABC would have two obtuse angles, a contradiction (in Euclidean geometry). Thus there is only one possible angle for B, the acute angle. We can now completely solve the triangle. Since a triangle is completely determined by SsA, this implies the SsA Congruence Theorem. QED
I don't remember the proof of SsA from the Third Edition of the text, but I know that it doesn't simply appeal to the Law of Sines, which isn't taught in any edition of the Geometry text. But it might give us a hint as to whether AaS is valid on the sphere or not.
We recall from Van Brummelen the past two summers that there is a spherical Law of Sines. In fact there are many trig formulas in that text. Most of them I have trouble remembering, but the spherical Law of Sines is one of the easiest to recall. It's just like the Euclidean Law of Sines, except that everything in the spherical formula requires the sine:
sin a/sin A = sin b/sin B
Now suppose A, B, and a are all given. We now solve for sin b:
sin b = (sin B sin a)/sin A
And while there are statements similar to Unequal Sides for spherical triangles, we realize that triangles can indeed have two "obtuse" sides (that is, longer than a quadrant). Thus knowing that both A > B and a > b doesn't allow us to rule out that b > 90. Thus AaS remains ambiguous.
Unfortunately, Van Brummelen has very little to say about AAS in his book. He writes more about both ASA and even SSA than AAS. I'm actually now starting to suspect that rather than AaS, it might actually be aAS (where the angle opposite the known side is the smaller angle) that can be fully determined on the sphere. If this is true, then my proof that requires AaS is surely invalid. Once again, I don't know whether there is a counterexample to our equiangular quad claim or not.
My Summer Plans for Spherical Geometry
Let me reveal what exactly my plans are for spherical geometry this summer. Note that not every post this summer will be about spherical geometry -- here I'm only referring to the spherical posts.
Each spherical post will be labeled as a "unit" -- as in a hypothetical unit of a high school course that begins with Euclidean geometry and ends with an intro to spherical geometry. This means that most of the posts will actually be about Euclidean geometry from another perspective.
The first such post will introduce our most important postulate -- Point-Line-Plane. In that post, I want to give the Euclidean (same as the U of Chicago), spherical, and natural versions. Once again, the "natural" Point-Line-Plane postulate is supposed to capture statements about points and lines that hold in both geometries.
In subsequent posts, we'll try to prove as many statements as possible using natural Point-Line-Plane as well as the other postulates of Euclidean geometry (including Playfair). We'll keep on emphasizing transformations just as in Common Core, including reflections and rotations. Translations might not necessary exist in spherical geometry.
Notice that in an actual class, we'd state only Euclidean Point-Line-Plane. I'd be careful to use only the parts of Point-Line-Plane that are natural, but we don't tell the students this. Only at the end when we introduce spherical geometry do we reveal which statements hold in both geometries.
At this point you might wonder, what the point of making these posts? For one thing, we know that spherical geometry was first created by Bernhard Riemann, yet very little writing from past two centuries since the Riemann sphere can be found online. My hope is that I can post and provide many of the proofs that I can't locate online.
Traditionalists: SteveH's June 15th Comment
I've added the "traditionalists" label to this post, since the traditionalists have been active lately.
In my last post, I linked to our favorite traditionalist, Barry Garelick, and his Flag Day post along with several comments. At the time, our favorite commenter SteveH hadn't posted yet. So let's discuss SteveH's comment right now:
https://traditionalmath.wordpress.com/2019/06/14/misunderstandings-about-understanding-dept/
SteveH:
Answer getting also requires a lot of understanding at the lowest levels of math, including the times table. What is 6*7? 42, isn’t it? Well 5*7=35 + 7 = 42, because 7 is 5+2 and I like to add by fives. Nobody does math only with rote memory. Then there are things like telling time and counting change with quarters, dimes, nickels, and pennies.
This is fine and dandy. Indeed, I like the idea of teaching 6 * 7 = 5 * 7 + 7 = 35 + 7 = 42. Indeed, we discussed this earlier on the blog regarding learning times tables in other bases. For example, since 6 is half of the dozenal base 12, multiplying by dozenal 7 is a bit like multiplying by decimal 6, and SteveH's trick here. A dozenalist would thus write 6 * 7 = 6 * 6 + 6 = 30 + 6 = 36 (dozenal).
But here's the problem. A third grade teacher teaches 6 * 7 = 5 * 7 + 7 = 35 + 7 = 42. A student has already memorized 6 * 7 = 42, and thus omits the "5 * 7 + 7 = 35 + 7 =" part. The teacher marks the student wrong, or deducts points, for omitting the "5 * 7 + 7 = 35 + 7 =" part. And the first to complain would be SteveH and the other traditionalists.
SteveH:
Most people forget the incremental understanding process they went through to get there. Telling time is very difficult, what with 365 days per year (plus leap year), 30 or so days per month (do we stop teaching kids the rote rhyme? I don’t even know the reason or understanding of the rhyme.), 24 hours per day (AM and PM), 60 minutes per hour, and 60 seconds per minute. Do they have to know about the solar system, years for different planets, spin on a tilted axis to the sun, and why 24 and 60 as bases? Do they have to know about different bases? My son had a thematic unit in first grade about little Eskimo kids in Alaska without ever knowing why it’s so cold and dark up there most of the year. Clearly, they aren’t worried about rote knowledge.
This is a tricky one. My priority is to address and avoid "Why?" questions asked by actual students sitting in actual classrooms. SteveH's first grade son probably never asked "Why is it so cold and dark up in Alaska most of the year?" -- thus the first grade teacher never had to explain it.
Actual students won't ask why there are 60 minutes per hour and 60 seconds per hour. Instead, they will ask "Why do we have to learn to tell time (on an analog clock)?" Many of them believe that the existence of digital time on cell phones renders learning analog time archaic. And some of them might even think that counting change with quarters, dimes, etc., is also archaic, because inflation has resulting in items costing more than a quarter these days -- also because they plan to do nearly 100% of all transactions with a credit or debit card anyway.
Some students might ask why 6 * 7 is 42. But others might ask why they need to learn their sixes and sevens times tables, since they believe that the existence of calculators renders memorizing the times tables archaic. What we wish to do is reach students who refuse to memorize that they believe that they shouldn't have to.
SteveH:
“Answer getting” is something they made up just to push their beliefs. So what’s driving it? All of that silliness disappeared (for the most part) when my son got to high school. It’s a K-8 thing and in our schools, the fundamental driving force in K-8 is full academic inclusion. Our high school, however, is full environmental inclusion. To get full academic inclusion to work, however, requires lowering grade level expectations and creating a world of magic natural differentiated learning where “kids will learn when they are ready.”
I've seen that phrase "full inclusion" mentioned by several traditionalists, not just SteveH. All I can glean from this posts is that "full inclusion" is bad. I'd think that "full inclusion" would be good because its negation, "someone is excluded" is bad. But to traditionalists, "full inclusion" is bad.
Here SteveH also distinguishes between "full academic inclusion" and "full environmental inclusion," with the former being definitely bad. I don't yet know (or understand) what the difference is. He mentions "full inclusion" once more in this post:
SteveH:
Full inclusion and CCSS have lowered the slope and expectations in K-8 and since the lowest level track in our high school is “College Prep”, then many students and parents are completely unprepared for the reality whack of honors and AP classes to STEM and what we called “College Prep” years ago.
We know that SteveH definitely considers the STEM track to be eighth grade Algebra I, freshman year Geometry, and senior year AP Calculus. He doesn't state what his ideal "College Prep" track would be. I know of at least one successful college grad who didn't take Geometry until his junior year -- and, as you'll soon see, his name is Barry Garelick.Traditionalists: Barry Garelick's June 19th Post
On Wednesday, Garelick himself posted again:
https://traditionalmath.wordpress.com/2019/06/19/my-first-educational-treatise/
My mother was my first critic of my first educational treatise, written when I was a senior in high school. During my junior year, I had tried unsuccessfully to get transferred out of a geometry class taught by a teacher whose reputation had preceded her for years.
Since this is a Geometry blog, I definitely want to read what Garelick has to say about Geometry.
In retrospect, she was a very bright woman who had emotional problems and spent the class talking about this, that and anything other than about geometry and left it to us to read the book, do the problems, and make presentations on the board. At that time, however, I wanted someone who taught.
Much of the debate regarding traditionalism vs. discovery math doesn't involve Geometry, but we definitely know of one Geometry text that's discovery -- Michael Serra's Discovering Geometry. I don't know whether Garelick's Geometry teacher used Serra's text, but we gather that based on the way she taught her class, it might as well be Discovering Geometry. She had her students discover geometric relations for themselves as opposed to teaching them the postulates and theorems.
In the end, Garelick wrote a pamphlet about how bad his teacher was, and even sold it:
I started selling the pamphlets at school for 25 cents a piece which got me in a little bit of trouble at school, but that’s another story. One day my mother got mad at me for something I said to her (probably complaining about something I didn’t like about something she did) and she let me have it. It was one of those “After all I’ve done for you” type rants, and once she got going there was no stopping her. Among the things she mentioned was running off the dittos and helping me glue them together.
See -- that's something you could have bought for a quarter back in the day. I wonder how much he would have sold it for if he were a young high school student nowadays.
Anyway, Garelick went on in the pamphlet to criticize English teachers and their grading policies. As this is a math blog, I don't comment on grades in English classes here.
He now moves forward to his teaching career, and tells a story about his previous middle school:
In that particular school, I was available for tutoring students prior to the first period, and some of her students would occasionally come in for help. In one case, a boy was having difficulty with proportion problems, such as 3/x = 2/5. I showed him how it is solved using cross multiplication (without explaining why cross multiplication works—I was after just getting him through the assignment). I ran into the boy’s father during the summer, and he made a point to thank me for helping his son. I found this odd since I had only tutored him maybe one other time. “He was getting D’s and F’s on his test and since you helped him, he started getting A’s and he ended up with an A- in the class.”
Proportions often come up in the traditionalists' posts. Since this story takes place in Garelick's past, it's possible that this was before the advent of Common Core. Nevertheless, it's likely that his colleague used Common Core methods of teaching proportions in middle school -- including double number lines and tape diagrams -- and these were confusing the student in the story.
Let's compare how Common Core teaches proportions in middle school to how it teaches another middle school standard, the Pythagorean Theorem. The Common Core developers figured that many students consider equations to be huge turnoff, and thus the idea was to show students how to solve as many problems as possible without needed to write an equation. This is why they use double number lines and tape diagrams to solve proportions. On the other hand, even the Common Core writers know that there is no easier way to solve a Pythagorean Theorem problem other than to use the underlying equation, a^2 + b^2 = c^2. That's why there is no special "Common Core method" of solving a right triangle problem.
Garelick tells us that this boy he tutored found it much easier to solve 3/x = 2/5 by cross multiplying than any method that the boy's own teacher showed him. But how many other students sitting in that boy's class are confused by cross multiplication? How many of them might try to write 5x = 6 or even 3x = 10 instead of the correct 2x = 15? The purpose of the double number lines and tape diagrams is so the student can visualize why x must clearly be greater than 5, so that 5x = 6 (leading to x = 1.2) can't possibly be correct?
Conclusion
Garelick clearly wants solving proportions by cross multiplying to be taught earlier than it currently is under the Common Core. He doesn't tell us how old the boy in his story was -- as far as we know, he was an Algebra I eighth grader. Let's give the other teacher the benefit of the doubt that he was a seventh grader.
I was actually thinking about when proportion equations should be taught while researching spherical vs. Euclidean geometry and the similarity proofs. We see how the U of Chicago Geometry text states theorems about slope without proof (since slope was taught in Algebra I) and then uses slope in coordinate proofs about S_k.
Perhaps the best plan to meet the Common Core Standards is to teach the derivation of slope from similarity in Algebra I. (The problem in eighth grade is that in many classes, all of the EE standards, including slope, are taught before any G standards, including similarity.) Then since similarity is being discussed in Algebra I anyway, that's when solving proportion equations by cross multiplying should definitely occur -- and be emphasized more than they currently are in Algebra I. Of course, that won't be good enough for Garelick and the traditionalists.
I look forward to writing our Point-Line-Plane postulates for spherical and natural geometry. But that probably won't be in my next post, since they will take some time for me to devise.
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