Monday, December 8, 2014

Section 7-5: The SSA Condition and HL Congruence (Day 75)

Today is Day 75. In the early start calendar, we are approaching the end of the first semester. In the Labor Day start calendar, we have ended the first trimester and are approaching the end of the third quaver -- that is, the 3/8 mark of the year.

Some readers who have noticed how I use the musical term quaver to denote an eighth of the year may have heard of a related musical word -- the hemidemisemiquaver or 64th note. As a musical note, this is much too fast to play by an ordinary pianist, or even a drummer. (I looked at a book of Christmas songs, and noticed that a vocal inflection in Mariah Carey's "All I Want for Christmas is You" was notated using near-64th notes.)

A hemidemisemiquaver of an academic year would be 1/64 of the year, or about three days. It is much too fast to correspond to any progress report, not even a weekly progress report.

The first Saturday in December is the day of the annual William Lowell Putnam competition. It is a math contest for college students. Ten years ago, a poster named John called it the "hardest math test in the world."

Every year, I like to inspire the students I tutor by showing them the first -- and usually easiest -- question, A-1, from this year's Putnam exam. I especially like to show this to my geometry students, since Putnam questions tend to be proofs. They may complain when they have to do proofs, but those proofs are nothing compared to Putnam proofs.

This year, question A-1 is a calculus question, but question B-1 is on digits -- something that my geometry student should understand. I'm scheduled to tutor him tomorrow, so I'll discuss this year's B-1 and the student response to it in tomorrow's post.

Section 7-5 of the U of Chicago text is on the final congruence theorem, HL, as well as why SSA doesn't work in general. I've been waiting to discuss all of the congruence theorems for a while now, and today's introduction of SSA and HL represent a good opportunity to discuss them.

Let's recall how our text proves SSS, SAS, and ASA. In each case, the proof in the text begins by having an isometry mapping one congruent side to its corresponding side in the other triangle. Now this isometry may already map the entire first triangle to the entire second triangle. If it doesn't, then the resulting figure formed by the second triangle and the image of the first is proved to be symmetrical (possibly because it forms a kite or isosceles triangle), which then implies that one more reflection maps the image of the first triangle onto the second triangle.

The book's proof of HL is similar. Here it is, converted to two-column form:

Given: BC = YZ, AB = XY, C and Z are both right angles
Prove: Triangles ABC and XYZ are congruent

Statements                      Reasons
1. BC = YZ, AB = XY,     1. Given
C and Z right angles
2. Map BC to YZ             2. Definition of congruent
3. ABC A'B'C' congruent 3. Definition of congruent
4. measure A'ZX = 180   4. Angle addition
5. AB = A'B'                    5. Definition of congruent
6. A'B' = XY                    6. Transitive Property of Congruence
7. A'YX isosceles             7. Definition of isosceles
8. angle A', X congruent  8. Isosceles Triangle Theorem
9. A'B'C' XYZ congruent 9. AAS Congruence Theorem
10. ABC XYZ congruent 10. Transitive Property of Congruence

Now let's compare this proof to those given by our other Common Core Geometry sources, Drs. Franklin Mason and Hung-Hsi Wu.

For SAS, Dr. M transforms the first triangle first by translating it until one pair of the vertices -- namely the pair where the congruent angles are -- coincide. Then a rotation is performed until one pair of congruent sides coincides. It could be the case that the triangles already coincide, and if not, then a single reflection will make them coincide. As it turns out, Euclid's "superposition" proof as mentioned in his Proposition 4 is similar to Dr. M's proof, except that Euclid forgot to consider the case where a reflection might be needed:

Dr. M's proof of SSS is a bit more complicated, and he leaves out some of the steps. But his proofs of ASA and HL are identical to the ones given by the U of Chicago. (The readers of this blog who don't have the U of Chicago text may wish to look at Dr. M's proofs.)

Reading the Dr. M and U of Chicago proofs make it clearer why SSS, SAS, and ASA work even in non-Euclidean geometry, while AAS and HL work only in Euclidean geometry, and SSA doesn't work at all. When one performs the isometry that maps one side to another, the other congruent sides and angles in SSS, SAS, and ASA are all near the reflecting line (that is, the side the second triangle and the image of the first have in common). But in AAS and SSA, one of the congruent angles is directly opposite the common side. So one can't use the Kite/Isosceles Triangle Symmetry Theorems or other tricks to conclude that the triangles are congruent. Of course, in AAS we use ASA plus the fact that the angle measure of a triangle is 180 degrees in Euclidean geometry. But there is nothing similar to that fact for side-lengths, so we still don't have SSA.

Dr. Wu gives proofs of SAS and SSS similar to Dr. M's. His proof of ASA (listed in the 8th grade section of his document) is similar to the SAS and SSS proofs. Actually, he considers the case where only a reflection is required and then gives what appears to be the U of Chicago/Dr. M proofs. In his Case 2, a rotation is required, and in his Case 3, a translation is required.

Notice that the U of Chicago text states what it calls the "SsA Congruence Theorem":

If, in two triangles, two sides and the angle opposite the longer of the two sides in one are congruent respectively to two sides and the angle opposite the corresponding side in the other, then the triangles are congruent.

We see that HL follows directly from SsA, since the congruent angles in HL are the right angles and they are always opposite the longest side of the triangle -- the hypotenuse. Had it not been for the fact that the SsA Theorem has a proof that is "quite difficult," the text would probably state HL as a corollary of HL.

In trig, we learn how SSA leads to the ambiguous case of the Law of Sines. In the SsA case, it turns out that one of the two possible triangles would have an angle of less than 0 degrees, which is why there is only one possible triangle. I wonder whether the "quite difficult" proof of HL that the book avoids is simply an appeal to the Law of Sines.

In explaining to my students why SSA is impossible, I take an isosceles trapezoid (that is not a rectangle) and cut it along its diagonal. We obtain two triangles that have two pairs of congruent sides and a pair of congruent angles, yet the triangles are not congruent -- that is, we get a counterexample to full SSA. I tell my students that if they think that SSA is a valid congruence theorem, they're an -- well, just spell SSA backwards to find out what they are...

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