Today is a special day, for those of you familiar with my blog over the years. It is my 38th birthday.
Also, today I subbed in a high school English class (different school from yesterday). There are two freshman English classes and an English as a Second Language (ESL) class -- a single class that lasts for two periods. There are students of all high school grades in this class.
I won't do "A Day in the Life" today, but I will describe the classes. All classes have a vocabulary quiz today -- ten words for ESL, twenty for the freshmen. All the words on each quiz start with the same letter -- "D" for ESL, "F" for the freshmen. (D and F -- those are the last letters I want to think about at quiz time!) In the second hour, the students work on Chromebooks (ESL, Studio 44). The last period of the day is team football. As I often do, I actually lift some weights with the athletes.
One thing I enjoyed back on my 36th birthday at the old charter school is the Conjectures game, also known as "Who Am I?" (I'd established this tradition the previous year, my 35th, when I was subbing in a math class.) And so I can't resist playing the game today as a review -- even though I only rarely played it in classes other than math. It also gives the students something to do, rather than just say "review your vocabulary for 20-25 minutes" (after which the students would talk or play games on their phones the whole time).
In second period ESL, the game doesn't work very well, probably because most of the students have very low English skills. Most of the students will apparently fail the quiz anyway. In the two freshman classes the game works out much better. The classes are a bit loud, but this game encourages it. This time I begin with "guess my college" (easy since I'm wearing a UCLA shirt). Then I do the usual "guess my weight" and of course, "guess my age" (since it's my birthday).
I name third period as the best class of the day, since it isn't quite as loud as fifth period. Third period is also more engaged with the quiz review. Recall that some questions in this game are answered as a "race" (first group to answer gets the point) while others require a response from every group. One of the latter type asks students to give the definition of "fawn." Some groups in third period answer "a spotted deer" while others say "yellowish-brown in color." Well, the original vocab worksheet lists "fawn" as an adjective, and so I only accept the color.
(Readers of this blog should already know that "fawn" is a color, since it's one of Kite's old colors in his notation for 19-limit music. And of course "Fawn" is also the name of one of my favorite math teacher bloggers, Fawn Nguyen.)
In the end, I like how my "Who Am I?" game often leads to a great discussion among the students -- even if it is a bit loud. Once again, this is why my game is much better suited to introducing new concepts rather than reviewing them before a test.
As it turns out, one student in this class is also celebrating his birthday today. He's a freshman, so this is his fifteenth birthday (just 23 years younger than I am). He's in the group that wins the "Who Am I?" game and so I write his name on my good list for the teacher -- and once again, he's in the best-behaved period of the day. But the birthday boy doesn't do well on his test, answering only three of the 20 vocab questions.
Some students believe that they're entitled to a non-academic free day on their birthday. I was never that student -- I always worked as hard as a young student on December 7th as I did any other day. But I recall one year when I was student teaching an Algebra II class. The master teacher and I decided to give the test the Tuesday before Thanksgiving (at the time, most schools in the area had classes through Wednesday of Thanksgiving week), since the next day was a minimum day and we figured that the students wouldn't concentrate on the test. But one girl refused to take the test because it was her birthday. In fact, she was perfectly willing to take the test on Wednesday, the day before the holiday, because it was no loner her birthday. And I think she passed the test with flying colors!
I don't ever remember having a major test on my birthday. But I do recall that in college, I had to take two major tests on December 8th (the GRE to get into grad school at UCLA, and two years later, the final exam for a set theory class). I have a funny story about the GRE. I knew that the GRE had math and verbal sections, but I didn't even know about the logic section until the night before the test (that is, my birthday night)! So I had to cram for an entire section on my birthday. Still, I know I must have passed that section because I obviously got into grad school at UCLA. (Note that the GRE logic section was eliminated the year after I took the test.)
Today, my 38th, is a milestone birthday. It is my second Metonic birthday, referring to the 19-year cycle of the moon. This means that the phase of the moon on my 19th, 38th, 57th, etc., birthdays should match the lunar phase on the day I was born. Today is a new moon, and so we conclude that I must have been born under a new moon as well.
(Notice that in my own Pacific Time Zone, the new moon occurred at 11:20 last night, but in every other time zone except Pacific, the new moon is today. If Proposition 7 were in effect, then the new moon would have been 12:20 AM on the 7th in California as well.)
New moons around this time of year occur during the Jewish holiday of Hanukkah. Observant Jews light the sixth Hanukkah candle tonight -- and typically, the new moon occurs around the sixth or seventh night of the holiday. Thus we conclude that the night I was born was probably also the sixth or seventh night of Hanukkah that year.
I actually remember learning about the Metonic cycle during my senior year in high school. I had obtained a free discarded book from our school library. It was a recreational math book that contained puzzles, including charts to help calculate the date of Easter. It was then that I learned about the 19-year cycle. I looked ahead on the calendar to the following December and noticed that my birthday would fall on a new moon. And this was my nineteenth birthday -- my first Metonic birthday. And so I realized that I must have been born under a new moon.
There's one more thing I want to say about 38th birthday. In one Simpsons episode, Homer remarks that since he is 38.1 years old and the average male lifespan is 76.2 years, he's exactly halfway through his life -- the definition of middle-aged. A Google search reveals that the average lifespan of an American male is now 78.6 years old, so I won't be middle-aged until next year, at age 39.3.
Lesson 7-5 of the U of Chicago text is called "The SSA Condition and HL Congruence." This lesson introduces the final congruence theorems.
This is what I wrote last year about today's lesson:
Lesson 7-5 of the U of Chicago text is on SSA and HL. I've already mentioned how I'll be able to prove HL without using AAS, since we have to wait before I can give an AAS proof.
Meanwhile, we know that SSA is invalid, but the U of Chicago text provides us with an SsA Congruence Theorem, where the size of the S's implies that it must be the longer of the congruent sides that is opposite the congruent angle. The text doesn't provide a proof of SsA because the proof is quite difficult -- certainly too difficult for high school Geometry students.
But you know how I am on this blog. I'm still curious as to what a proof of SsA entails, even if we don't ask high school students to prove it. Last year, I mentioned how SsA leads to the ambiguous case of the Law of Sines.
I've noticed that when using the Law of Sines to solve "both" triangles for the SsA case, the "second" triangle ends up having an angle sum of greater than 180 degrees. We can use the Unequal Sides Theorem to see why this always occurs -- that theorem tells us that the angle opposite the "s" must be smaller than the angle opposite the "S" (the known angle). We know that if two distinct angles between 0 and 180 have the same sine, then they are supplementary. So the second triangle would have angle sum of at least 180 minus the smaller angle plus the larger angle -- which always must be greater than 180. (This is often called the Saccheri-Legendre Theorem, named for two mathematicians with whom we're already familiar and associate with non-Euclidean geometry. Of course spherical geometry is not neutral, and Saccheri-Legendre fails in spherical geometry.)
That the sum of the angles of a triangle can never be greater than 180 (but we don't necessarily know that it's exactly 180) is neutral, but the Law of Sines is not neutral. Nonetheless, it is known that SsA is a neutral theorem. So the Law of Sines can't be behind the secret proof that U of Chicago text doesn't print in its text.
But one of the questions from the U of Chicago text hints at how to prove SsA -- and it's one that I included on my HL/SsA worksheet last year. Here is the question:
Follow the steps to make a single drawing of a triangle given the SSA condition:
a. Draw a ray XY.
b. Draw angle ZXY with measure 50 and XZ = 11 cm.
c. Draw circle Z with radius 9 cm. Let W be a point where circle Z and ray XY intersect.
d. Consider triangle XZW. Will everyone else's triangle be congruent to yours?
The answer is that they most likely won't. In step (c) we are to let W be a point -- not the point -- where the circle and the ray intersect. This implies that there could be more than one point where they intersect -- and in fact, there are two such points. But there can never be a third such point, because a circle and a ray (or line) intersect in at most two points. We can prove this indirectly using the Converse of the Perpendicular Bisector Theorem:
Lemma:
A line and a circle intersect in at most two points.
Indirect Proof:
Assume towards a contradiction that there exists a circle O that intersects line l in at least three points A, B, and C, and without loss of generality, let's say that B is between A and C. By the definition of circle, O is equidistant from A, B, and C. From the Converse of the Perpendicular Bisector Theorem, since O is equidistant from A and B, O lies on m, the perpendicular bisector of
Now both m and n are said to be perpendicular to l, since each is the perpendicular bisector of a segment of l. So by the Two Perpendiculars Theorem, m and n must be parallel -- and yet O is known to lie on both lines, a blatant contradiction. Therefore a line and a circle can't intersect in three points, so the most number of points of intersection is two. QED
So let's prove SsA now using this lemma and the construction from the problem above. Once again, the proof must be indirect. Even though most of our congruence theorem proofs call the two triangles ABC and DEF, I will continue to use the letters XZW so that it matches the above question.
Given: AB = XZ < BC = ZW, Angle A = Angle X
Prove: Triangles ABC and XZW are congruent.
Proof:
We begin by performing the usual isometry that maps AB to XZ. As usual, we wish to show that the final reflection over line XZ must map C to W -- that is, C' must be W.
So assume towards a contradiction that C' is not W. As usual, the given pair of congruent angles allows us to use the Flip-Flop Theorem to map ray AC to ray XY. Just as in the problem from the text, we know that C' must be a point on ray XY (since ray AC maps to ray XY), and it must be the correct distance from Z (since reflections preserve distance). Now the set of all points that are the correct distance from Z is the circle mentioned in the above problem. We know that the intersection of a circle and a ray is at most two points, and so the assumption that C' is not W implies that W must be one of the two points of intersection, and C' must be the other point.
We must show that this leads to a contradiction in the SsA case -- that is, when AB and XZ are longer than the sides BC and ZW. We see that if C' and W are distinct points equidistant from Z, then the triangle ZC'W must be isosceles, and so its base angles ZC'W and ZWC' must be congruent.
Now we look at triangle ZC'X. It contains an angle, ZC'X, which forms a linear pair with ZC'W, so its measure must be 180 - m/ZC'W -- that is, m/ZC'X = 180 - m/ZWC' by substitution. It also contains an angle ZXC' -- which (renamed as ZXW), we see must be larger than ZWC' (renamed as ZWX) by the Unequal Sides Theorem -- ZXW is opposite the longer side ZW (in triangle ZXW), so it must be the bigger angle.
So now we add up the measures of two of the three angles in triangle ZC'W -- the angle ZC'X has measure 180 - m/ZWC', and angle ZXC' is known to be greater than ZWC'. So the sum of the two angles is greater than 180 - m/ZWC' + m/ZWC' -- that is, it is greater than 180. That is, the sum of the angles of triangle ZC'W is greater than 180 -- which is a contradiction, since by the Triangle-Sum (actually Saccheri-Legendre) Theorem, the sum must be at most 180.
Therefore the assumption that C' and W are different points is false. So C' must be exactly W. QED
Of course we wouldn't want to torture our students with this proof. It depends on three theorems -- Isosceles Triangle, Unequal Sides, and Triangle-Sum -- that we have yet to prove. On my worksheet, I added an extra note to this problem explain how it leads to SsA.
Again, I retain references to non-Euclidean geometry from the old post. The new Third Edition of the U of Chicago text includes an actual SsA proof. It's similar, but not quite like, the proof given here. I dropped the mention of SsA on the worksheet since today's the second day of the SSASS activity. Oh, and since SsA works, so does SsAsS. Two short sides are adjacent to the angle, which implies a convex quadrilateral.
Oh, and since the Spherical Geometry label is on this post, it makes me wonder about the Quadrilateral Congruence Theorems in Spherical Geometry. I'm not quite sure which theorems are spherically valid, since very few authors write about quadrilateral congruence on the spherical. I suspect that at least SASAS and ASASA would be valid on the sphere.
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