Tuesday, February 2, 2016

Lesson 12-8: The AA and SSS Similarity Theorems (Day 95)

Lecture 21 of David Kung's Mind-Bending Math is "More With Less, Something for Nothing." In this lecture, Kung introduces the concept of geometric optimization -- doing the most with the least.

Kung begins with a classic problem -- suppose from your campsite you see a fire. What's the shortest path from your campsite to a river (to get a bucket of water) to the fire? The answer, as Kung points out, involves reflections. If the riverbank is at r, the campsite is at C, and the fire is at F, then we reflect F across r to obtain F'. Then the shortest path is the segment CF'. I've seen this problem mentioned in other Geometry texts, so I'm surprised that it doesn't appear in the U of Chicago text.

Kung adds that it's possible to take differing speeds into account -- for example, it may be slower to run with a bucket full of water than it is with an empty bucket. He explains that the solution is similar to Snell's Law for optics, when the speed of light is different in air and water. (Einstein's Law -- that the speed of light never changes -- only applies to speed in a vacuum.)

The next set of problems Kung mentions are called isometric problems. Of all rectangles with a fixed perimeter, which one has the greatest area? The answer is a square -- and he points out that this can be proved using either Algebra II or Calculus. If we're allowed to consider any plane figure -- not just rectangles -- then the greatest area is the circle. This is actually mentioned in the U of Chicago text, as is the 3D analog -- of all solids with a given surface area, the largest volume is the sphere. The Isoperimetric Theorems are given in the last two lessons of the text, Lessons 15-8 and 15-9, but we didn't cover either of these last year.

So do I seem to be always skipping the lessons that are the most fun? Well, last year I was running out of time to cover Chapter 15 before the PARCC, and I knew that nothing after Lesson 15-3 actually appears on any Common Core test.

Here's the Quick Conundrum, and it's related to the Isometric Theorems -- two balloons, one fully inflated, the other less inflated. Kung sets them up so that air can flow freely from one to the other -- so which way will the air flow? As it turns out, air flows from the less inflated balloon to the more inflated balloon -- because smaller balloons actually have more pressure. (This is Boyle's Law for ideal gases.)

Kung moves on to Steiner trees, named for the 19th century Swiss mathematician Jakob Steiner. We have three cities A, B, C, that we wish to connect with this shortest roads possible. So we choose a point P and connect the three roads PA, PB, PC. So where must point P be? Considering last month's lessons here on the blog, four guesses immediately come to mind:

-- the centroid of Triangle ABC
-- the circumcenter of Triangle ABC
-- the incenter of Triangle ABC
-- the orthocenter of Triangle ABC

As it turns out, none of these are correct. The answer is the Fermat point of Triangle ABC -- named for the 17th century French mathematician Pierre de Fermat. The Fermat point P is chosen so that the angles APB, BPC, and CPA are all 120 degrees. Here's how to construct a Fermat point -- begin by constructing three equilateral triangles, outside of BCCA, and AB. Let's label the three new triangles A'BC, AB'C, and ABC'. The three segments AA', BB', and CC' are -- you guessed it -- concurrent, and they intersect at the Fermat point P. (The Fermat point is easy to construct after completing Level 1 of Euclid the Game -- since this constructs an equilateral triangle.)

Kung ends his lesson by discussing the Kakeya Problem -- named for a Japanese mathematician, Soichi Kakeya. We ask, how much area does it take to turn a needle? That is, if you take a line segment that is 1 unit long and turn it counterclockwise, how much space do you need?

This question is obviously related to Common Core rotations, since we're rotating a needle. Kung writes that the naive answer pi/4, obtained by rotating the segment about its midpoint, is not even close to correct -- we can do much better.

The solution, given two years later by Russian mathematician Abram Besicovitch, actually involves translations as well as rotations. Note that all isometries of the plane are either a rotation, translation, reflection, or glide reflection -- if we take two congruent triangles and toss them in the air so that they land on the plane of the floor, one of these four isometries must map one to the other. And if we know that the two triangles have the same orientation, then a translation or rotation must be correct.

If we take two needles pointing in the same direction, then a translation maps one to the other. If they are pointing in different directions, then there must exist a rotation mapping one to the other -- though the center of that rotation may be far away, if the degree difference is small.

This works in reverse. If we take a needle, translate it in the direction it is pointing, rotate it by a small angle around its midpoint, then translate it the same distance but opposite the direction it is now pointing, then this is equivalent to a simple rotation of the original needle (same center and rotation), but using much less area. After all, as Kung points out, translating the needle in the direction it's pointing uses no area, and the rotation is small, so it takes very little area. If the translation part is done correctly, the area over which each tiny rotation occurs can overlap. Kung states that the final sequence makes the needle rotate over an arbitrarily small area.

Let's get to our Geometry lesson. We are now working on the AA~ and SSS~ theorems, which complete our study of similarity. There are several ways we can prove these at this point. We can use the original dilation proofs given in the U of Chicago text (Lessons 12-8 and 12-9), or we can use the one similarity theorem we already have (SAS~) plus the corresponding congruence theorems (ASA and SSS, respectively).

Kung wraps up the lecture by hinting that if we wanted just to have just a line segment pointing in every direction (without rotating anything), the minimum area is actually zero. But what figure can have zero area?

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