1. Introduction
2. Pappas Problem of the Day
3. The Reflection Postulate
4. Line Perpendicular to Mirror Theorem
5. Line Parallel to Mirror Theorem
6. SSS, SAS, and ASA Congruence Theorems
7. Old Charter School Update
8. Conclusion
Introduction
So far, the natural geometry we learned in yesterday's post corresponds to Lessons 1-1 through 3-3 of the U of Chicago text. In that post, there was a Natural Point-Line-Plane Postulate. And since the Triangle Inequality of Lesson 1-9 also holds on the sphere, we can keep it as a postulate. The Angle Measure Postulate remains since of course we want to measure angles on the sphere (and possibly even use angle measures to define distance on the sphere). This means that the first unnatural (that is, spherically invalid) postulate is in Lesson 3-4:
Corresponding Angles Postulate:
If two coplanar lines are cut by a transversal so that corresponding lines have the same measure, then the lines are parallel.
We know that this is spherically invalid since it's definitely possible for corresponding angles to be congruent, and yet there are no parallel lines on the sphere.
Our goal is to keep our geometry natural as long as possible until we're forced to accept an unnatural postulate to develop Euclidean geometry. In several past posts, I suggested naming the units of a Common Core Geometry course after the transformations. Chapter 4 of the U of Chicago text is also about reflections. Since reflections also exist on the sphere, we can jump directly into Chapter 4, and so our second unit is titled "Reflections."
Pappas Problem of the Day
Today on her Mathematics Calendar 2019, Theoni Pappas writes:
If AB = 2/3, determine the perimeter of this regular hexagon.
(Here is the given info from the diagram: A is a vertex of the hexagon. To find B, consider the two vertices adjacent to A. Pappas doesn't label these vertices, so we'll call them X and Y. Then B is the point on ray XA such that ray XA is perpendicular to
We see that Angle YAB is an exterior angle of a regular hexagon, so it's 360/6 = 60 degrees. This makes Triangle YAB a 30-60-90 triangle, with 2/3 as its shorter leg. Thus the hypotenuse of this triangle is YA = 4/3 -- which is also a side of the regular hexagon. We multiply 4/3 * 6 = 8. Therefore the perimeter of the regular hexagon is 8 -- and of course, today's date is the eighth.
(Oh, let's not leave tau out of this post! The sum of the angles of a convex polygon is just tau, and so the sum of the angles of a regular hexagon is tau/6. This is another result that's easy to find using tau.)
Of course, we solved today's problem using Euclidean geometry. The sum of the exterior angles of a spherical polygon is less than 360, and hence there are no 30-60-90 triangles on the sphere.
(It might be interesting if somehow "two wrongs make a right" on the sphere -- even though the exterior angles of a regular hexagon are less than 360 and so YAB isn't 30-60-90, it might nonetheless hold that YA = 2AB. I seriously doubt this is the case -- but if it did, we would have stumbled upon a nice result of natural geometry. But this problem isn't the topic of today's post.)
The Reflection Postulate
Here is the definition of "reflection" as given in Lesson 4-1 of the U of Chicago text:
- For a point P not on a line m, the reflection image of P over line m is the point Q if and only if m is the perpendicular bisector of
PQ. - For a point P on m, the reflection image of P over line m is P itself.
Under a reflection:
a. There is a 1-1 correspondence between points and their images.
b. If three points are collinear, then their images are collinear.
c. If B is between A and C, then the image of B is between the images of A and C.
d. The distance between two preimages equals the distance between their images.
e. The image of an angle is an angle of the same measure.
For the most part, this definition and postulate applies equally on the sphere. But there may be a few concerns here.
First of all, "between" might be difficult to define on a sphere since all lines are circles. I'm not sure whether it's truly necessary to define "between" on the sphere. If we wish to have a definition, we might say that B is between A and C when B lies on the minor arc AC.
I'm not quite sure which points lie between N and S if these points are antipodal. The only choices that make sense are "no points lie between N and S" and "all other points lie between N and S. "The former allows us to say "if both B and C lie between A and D, then all four are collinear," which also holds in Euclidean geometry.
Closely related to "betweenness" is the idea of a "ray." Rays can be defined in terms of betweenness:
- The ray with endpoint A and containing a second point B consists of the points on
ABand all points which B is between each of them and A.
Rays are significant mainly because angles are defined in terms of rays:
- An angle is the union of two rays that have the same endpoint.
And the Angle Measure Postulate (or Protractor Postulate) associates rays with numbers from 0-180, so our definition of rays allows angles to be defined as well. Notice that angles consist of rays and triangles contain angles, and yet triangles don't contain rays -- this is true in both Euclidean and spherical geometry. The angles of a triangle appear when we extend their sides to be rays -- either semicircles on the sphere, or infinitely on the Euclidean plane.
But there's one more problem with reflections on the sphere -- antipodal points. If N and S are antipodal, does there exist a reflection mapping N to S? After all, if segment NS is undefined, how can it have the perpendicular bisector mentioned in the definition.
There are infinitely many semicircles from N to S. Each one of these has its own bisector, a point that lies a quadrant away from either pole. But it's not obvious that all of these bisectors are collinear -- much less that this common line is perpendicular to all of the semicircles from N to S. Of course, we know that this is indeed the case, and we call that line the Equator -- but how can we prove it?
Implicit in the Reflection Postulate is the statement that if N and S are two distinct points, then there exists a unique reflection mapping N to S. This is true for both Euclid and the sphere -- including antipodal points on the sphere. If we wish, we can directly state this in a natural Reflection Postulate.
Then this immediately forces the Equator to exist and be perpendicular to every meridian. Indeed, we see that every pair of antipodal points has its own equator, and every line has its own poles. We saw how Legendre and Van Brummelen regularly referred to poles of a line and equator of a pair of antipodal points in their respective spherical texts.
Of course, in our high school Geometry class we don't mention the special antipodal case -- since all we're doing is stating the Reflection Postulate as given in the text and assuming that there exists a unique mirror mapping any given point to any other given point.
Line Perpendicular to Mirror Theorem
All of the remaining theorems of Chapter 4 hold in both Euclidean and spherical geometry -- that is, they are all natural. These are Figure Reflection Theorem, Perpendicular Bisector Theorem, Flip-Flop Theorem, Segment Symmetry Theorem, Side-Switching Theorem, and Angle Symmetry Theorem.
There are two theorems which I've found useful on the blog, but aren't stated in the text. The first is:
Line Perpendicular to Mirror Theorem:
If l is perpendicular to m, then the reflection image of l over m is l itself.
This theorem is simply the application of the Side-Switching and Angle Symmetry Theorems to a straight (180-degree) angle. In this case, any point on the line can be taken as the vertex of that straight angle. It holds in both Euclidean and spherical geometry.
This is where I often uses the words "fixed point" and "invariant line" of a transformation. A fixed point of a transformation T is a point P such that T(P) = P. According to the definition of reflection, every point on the mirror of the reflection is a fixed point.
An "invariant line" of a transformation T is a line l such that T(l) = l. The points on l are mapped to points on l, though not necessarily to themselves. The theorem above tells us that every line perpendicular to the mirror is invariant. The mirror itself is also an invariant line -- but not only is it invariant, but every single point on it is a fixed point.
Line Parallel to Mirror Theorem
Here is the other useful theorem:
Here is the other useful theorem:
Line Parallel to Mirror Theorem:
If l is parallel to m, then the reflection image of l over m is also parallel to m.
This theorem holds vacuously in spherical geometry, since there are no parallel lines. There is a closely related theorem that isn't vacuous in spherical geometry:
Line Intersecting Mirror Theorem:
A line l intersecting its reflection image over m at a single point (or at two points) must also intersect mirror m itself at that point (or two points).
This statement holds in both Euclidean and spherical geometry, but proof is tricky. One direction is trivial -- if l intersects m at point P, then l', the reflection of l, must pass through the reflection of P -- which is P itself, since it lies on the mirror m.
But the converse requires more work. If l intersects l' at point P, then P, as a point on the image line l', must be the image of some point on l -- but that point might not be P itself. It's possible for Q on l, a distinct point from P, could be the point whose image is P on l'. But then by the Flip-Flop Theorem, the image of P on l must be Q on l'.
Now in Euclidean geometry, we'd already be done. Since both P and Q lie on both l and l', this forces l and l' to be the same line. Thus l is invariant, and so it no longer intersects its image at a single point, which is a contradiction.
For spherical geometry, we must now consider the two-point case. Then the line and its image intersect at antipodal points. But then the mirror m must be the equator of those two points -- and any line through the antipodal points must be perpendicular to that equator. This forces l to be invariant, which leads to the same contradiction.
I'm still wondering whether this has a more elegant proof. The theorem shows that if a line intersects its image, than that line, the image, and the mirror are concurrent. This might turn out to be important in spherical geometry. (I suspect that this looks more elegant in "single elliptic geometry," where antipodal points count as one point. Then we don't need to mention a two-point case.)
SSS, SAS, and ASA Congruence Theorems
These theorems can be shown using the same proofs as in Chapter 6. They only require reflections, and so these are valid on both the Euclidean plane and the sphere.
Here's what a proof sketch of SAS might look like:
Given: AB = DE, AC = DF, Angle A = D
Prove: Triangle ABC = DEF
Proof sketch:
The first reflection maps A to D, so the first image of ABC is DB'C'.
The second reflection mapsDB' to DE since it's the correct length, so the second image is DEC".
Now since Angle EDC" is the correct size, ray DC" must either already be ray DF, or else the third reflection can map it to DF. Since DC" (or DC'") is the correct length, C" (or C'") must be F. Thus the final image of ABC is DEF. QED
In some Common Core texts, we often see a translation mapping A to D and a rotation mapping the segmentsDB' to DE. But we can do it all using reflections.
Old Charter School Update
I drove to several places today. (One of those places just happened to be near Northridge -- the epicenter of the last major earthquake before last week's.)
One of those places is my old charter school. The place looks the same as it did a few weeks ago, including the old charter signs -- except that there is a "Clutter" truck and worker there. "Clutter" is a moving and storage company.
Afterwards, I decide to drive past our sister charter school. This time, there are no charter signs -- and there's hardly any sign that the sister charter has ever been there. I think there might be a door near the interior with our logo on it, but that's it.
Now I'm wondering whether it's possible that our administrators are trying to appeal to the county or state, but with only one site -- our own -- rather than both ours and our sister charter's site. Recall that the sister phone number leads directly to an error message, while our phone number at least leads to a voice message (that no one responds to).
I'll try driving past my old school site again soon. By then the "Clutter" worker might have finished his work, and so I'll check to see whether any of our signs are still there.
I'm still thinking about my old seventh grade class -- the class that, as I explained in an earlier post, was my most important. I still can't help but think that, if only I'd done a better job teaching my seventh graders, I'd still be working there and both campuses would still be open.
One of my most successful seventh grade classes occurred that year on February 2nd. (Yes, it would have been nice if, like Groundhog Day, I could have repeated that great day over and over.) I know that the February 2nd class went well because, as I explained in my February 3rd post the next day, the seventh graders excelled on their tests.
One reason for success in this class might have been the way my traditional lesson was set up:
CCSS.MATH.CONTENT.7.EE.A.1
Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.
Even though this lesson was based on "rational" (read: integer) coefficients, most of my problems had only positive coefficients. This allowed students who struggled with negative numbers to be nonetheless successful during this lesson. Then I introduced negatives the following week.
So I wouldn't want to change this lesson much at all. If we compare this to the lesson plan I described in my June 30th post, the week of January 30th-February 3rd should indeed be a math test week. But then Thursday, February 2nd, should have been a science project. Notice that earlier that week on the last day of January, I gave a STEM project on the heart rates of different animals. Since seventh grade science should indeed have been life science, it's possible that I might have assigned a similar project out of the life science text.
Of course, not everything was rosy that week, as sixth grade struggled with their test that week. But once again, it shows how I could have been more consistently successful with the seventh graders.
Meanwhile, I've also been thinking about how I scheduled my Dren Quizzes. In my June 30th post, I wrote about how Dren Quizzes should be every four weeks, 10's and then 2's through 9's. I wonder whether I should have given multiplication review worksheets much more often. (Recall that my counterpart at the sister charter school regularly gave her students multiplication practice.)
Conclusion
I'm still trying to decide how far Unit 2 on reflections should go. If we follow the naive order that is suggested by the U of Chicago text, then we should let reflections cover Chapters 4 and 5, followed by the translations unit covering Lesson 6-2 (and then going back to parallel lines from Chapter 3 -- natural geometry ends here) and then the rotations taking us from Lesson 6-3 through Chapter 7. But in this post, I already hinted that SSS, SAS, and ASA can occur under reflections.
I like the idea of including, in the reflections unit, figures with reflection symmetry. This means that not only should isosceles triangles appear here, but also kites and isosceles trapezoids. Both of these have close analogs in spherical geometry (with two pairs of adjacent congruent angles, instead of isosceles trapezoids, as I explained yesterday).
On the other hand, we might include all triangles in one unit and quadrilaterals in another. I like the idea of all triangles under reflections, and since triangles have three sides, maybe that should be called Unit 3 instead of Unit 2 (forcing me to split Unit 1 in half).
Well, that's another decision I must make before my next post.
This statement holds in both Euclidean and spherical geometry, but proof is tricky. One direction is trivial -- if l intersects m at point P, then l', the reflection of l, must pass through the reflection of P -- which is P itself, since it lies on the mirror m.
But the converse requires more work. If l intersects l' at point P, then P, as a point on the image line l', must be the image of some point on l -- but that point might not be P itself. It's possible for Q on l, a distinct point from P, could be the point whose image is P on l'. But then by the Flip-Flop Theorem, the image of P on l must be Q on l'.
Now in Euclidean geometry, we'd already be done. Since both P and Q lie on both l and l', this forces l and l' to be the same line. Thus l is invariant, and so it no longer intersects its image at a single point, which is a contradiction.
For spherical geometry, we must now consider the two-point case. Then the line and its image intersect at antipodal points. But then the mirror m must be the equator of those two points -- and any line through the antipodal points must be perpendicular to that equator. This forces l to be invariant, which leads to the same contradiction.
I'm still wondering whether this has a more elegant proof. The theorem shows that if a line intersects its image, than that line, the image, and the mirror are concurrent. This might turn out to be important in spherical geometry. (I suspect that this looks more elegant in "single elliptic geometry," where antipodal points count as one point. Then we don't need to mention a two-point case.)
SSS, SAS, and ASA Congruence Theorems
These theorems can be shown using the same proofs as in Chapter 6. They only require reflections, and so these are valid on both the Euclidean plane and the sphere.
Here's what a proof sketch of SAS might look like:
Given: AB = DE, AC = DF, Angle A = D
Prove: Triangle ABC = DEF
Proof sketch:
The first reflection maps A to D, so the first image of ABC is DB'C'.
The second reflection maps
Now since Angle EDC" is the correct size, ray DC" must either already be ray DF, or else the third reflection can map it to DF. Since DC" (or DC'") is the correct length, C" (or C'") must be F. Thus the final image of ABC is DEF. QED
In some Common Core texts, we often see a translation mapping A to D and a rotation mapping the segments
Old Charter School Update
I drove to several places today. (One of those places just happened to be near Northridge -- the epicenter of the last major earthquake before last week's.)
One of those places is my old charter school. The place looks the same as it did a few weeks ago, including the old charter signs -- except that there is a "Clutter" truck and worker there. "Clutter" is a moving and storage company.
Afterwards, I decide to drive past our sister charter school. This time, there are no charter signs -- and there's hardly any sign that the sister charter has ever been there. I think there might be a door near the interior with our logo on it, but that's it.
Now I'm wondering whether it's possible that our administrators are trying to appeal to the county or state, but with only one site -- our own -- rather than both ours and our sister charter's site. Recall that the sister phone number leads directly to an error message, while our phone number at least leads to a voice message (that no one responds to).
I'll try driving past my old school site again soon. By then the "Clutter" worker might have finished his work, and so I'll check to see whether any of our signs are still there.
I'm still thinking about my old seventh grade class -- the class that, as I explained in an earlier post, was my most important. I still can't help but think that, if only I'd done a better job teaching my seventh graders, I'd still be working there and both campuses would still be open.
One of my most successful seventh grade classes occurred that year on February 2nd. (Yes, it would have been nice if, like Groundhog Day, I could have repeated that great day over and over.) I know that the February 2nd class went well because, as I explained in my February 3rd post the next day, the seventh graders excelled on their tests.
One reason for success in this class might have been the way my traditional lesson was set up:
CCSS.MATH.CONTENT.7.EE.A.1
Apply properties of operations as strategies to add, subtract, factor, and expand linear expressions with rational coefficients.
Even though this lesson was based on "rational" (read: integer) coefficients, most of my problems had only positive coefficients. This allowed students who struggled with negative numbers to be nonetheless successful during this lesson. Then I introduced negatives the following week.
So I wouldn't want to change this lesson much at all. If we compare this to the lesson plan I described in my June 30th post, the week of January 30th-February 3rd should indeed be a math test week. But then Thursday, February 2nd, should have been a science project. Notice that earlier that week on the last day of January, I gave a STEM project on the heart rates of different animals. Since seventh grade science should indeed have been life science, it's possible that I might have assigned a similar project out of the life science text.
Of course, not everything was rosy that week, as sixth grade struggled with their test that week. But once again, it shows how I could have been more consistently successful with the seventh graders.
Meanwhile, I've also been thinking about how I scheduled my Dren Quizzes. In my June 30th post, I wrote about how Dren Quizzes should be every four weeks, 10's and then 2's through 9's. I wonder whether I should have given multiplication review worksheets much more often. (Recall that my counterpart at the sister charter school regularly gave her students multiplication practice.)
Conclusion
I'm still trying to decide how far Unit 2 on reflections should go. If we follow the naive order that is suggested by the U of Chicago text, then we should let reflections cover Chapters 4 and 5, followed by the translations unit covering Lesson 6-2 (and then going back to parallel lines from Chapter 3 -- natural geometry ends here) and then the rotations taking us from Lesson 6-3 through Chapter 7. But in this post, I already hinted that SSS, SAS, and ASA can occur under reflections.
I like the idea of including, in the reflections unit, figures with reflection symmetry. This means that not only should isosceles triangles appear here, but also kites and isosceles trapezoids. Both of these have close analogs in spherical geometry (with two pairs of adjacent congruent angles, instead of isosceles trapezoids, as I explained yesterday).
On the other hand, we might include all triangles in one unit and quadrilaterals in another. I like the idea of all triangles under reflections, and since triangles have three sides, maybe that should be called Unit 3 instead of Unit 2 (forcing me to split Unit 1 in half).
Well, that's another decision I must make before my next post.
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