Thursday, March 7, 2019

Lesson 12-5: Similar Figures (Day 125)

I feel that since this is a math blog, it's my duty to comment on Google Doodles for STEM. Well, today's featured Doodle is definitely STEM-related.

Olga Ladyzhenskaya was a twentieth century Russian mathematician. She was famous for finding solutions to the Navier-Stokes equations for fluid flow.

Let's go to my favorite website for reading the biographies of mathematicians:

http://www-history.mcs.st-andrews.ac.uk/Biographies/Ladyzhenskaya.html

As we've seen before with other female mathematicians, the young Olga first became interested in math because she was the daughter of a mathematician:

He started teaching his daughters mathematics in the summer of 1930 beginning with giving explanations of the basic notions of geometry, then he formulated a theorem and in turn made his daughters prove it. It became apparent that Olga showed a strong talent for logical thinking from an early age.

And of course this is a Geometry blog, so we can appreciate the young Olga's interest in Geometry and her budding talent for logic. But unfortunately, the Soviet regime interrupted her studies:

However, this did not stop her father inspiring his pupils and his daughters. Olga's two sisters were forbidden to finish their studies, being expelled from school, but the authorities allowed Olga to finish her studies. However, Olga had problems continuing her education since she was the daughter of an "enemy of the nation". When she was fifteen years old, in 1937, her father was arrested by Stalinist authorities and executed without trial.

And yes -- we've read about similar tragedies involving young Jewish mathematicians who grew up in Germany before World War II. Fortunately, the young Olga is ultimately admitted to college:

In 1943 she became a student at Moscow State University (MGU) due to the intervention of the mother of one of her pupils who, on returning to Moscow, persuaded the rector to invite Olga to MGU. It was not easy for her to leave her teaching post and there were many battles with the school authorities before she could become a student. At University Olga's love of mathematics blossomed and she was awarded a Stalin stipend and a labourers ration card without which she would have been unable to survive. It was here where she first started studying algebra, number theory and subsequently partial differential equations.

By the way, Olga was eventually nominated for a Fields Medal (which I discussed in several previous posts), but she didn't win it. You can read the rest of her fascinating biography at the above link.

OK, that's enough about Ladyzhenskaya. Let's get back to similarity.

This is what I wrote last year about today's lesson:

Lesson 12-5 of the U of Chicago text is about similar figures. There is not much for us to change about this lesson from last year, except for the definition of similar itself. Recall the two definitions:
  • Two polygons are similar if corresponding angles are congruent and sides are proportional.
  • Two figures are similar if there exists a similarity transformation mapping one to the other.
The first definition is pre-Core, while the second is Common Core. The U of Chicago text, of course, uses the second definition. But that PARCC question I mentioned last week must be using the first definition, since it requires that we know what similar means before we can define dilations and ultimately similarity transformations.

In the U of Chicago text, the Similar Figures Theorem is essentially the statement that the second definition implies the first definition. We would actually need to prove the converse -- that the first definition (at least for polygons) implies the second. But the proof isn't that much different -- suppose we have two figures F and G satisfying the first definition of similar -- that is, corresponding angles are congruent and sides are proportional, say with scale factor k. Then use any dilation with scale factor k to map F to its image F'. Now F' and G have all corresponding parts congruent, so there must exist some isometry mapping F' to G. Therefore the composite of a dilation and an isometry -- that is, a similarity transformation -- maps F to G. QED


Actually, there is one thing to change this year. Today's Thursday, which means that I must remove last year's activity and replace it with exercises. I've been making wholesale changes to most of the worksheets this week, but today the only change I need to make is removing the activity.


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