Wednesday, December 19, 2018

Semester 1 Final Exam (Day 83)

Today I subbed in a high school English class. In fact, I just subbed in this class two weeks ago, back on December 6th. Thus I wrote about this class back in my December 6th post.

On that day I didn't write "A Day in the Life," and I won't do so today either. But if you recall from two weeks ago, this teacher has only three classes -- one junior class and two senior classes. The juniors today are reading and annotating the short story The Devil and Tom Walker. I introduce myself to them as Mr. Walker -- and inform them that no, my name isn't Tom.

Two weeks ago, the regular teacher delayed a Siddhartha quiz that the seniors were supposed to take, but today they're required to take a test on the novel. Afterwards, they watch part of the movie.

The desks are normally organized in groups, but the regular teacher tells me to have the students put them in rows for the test. I tell fourth period to do so after taking attendance, and I have keep them in rows for fifth period. This means that I can't use the seating chart for fifth period attendance -- and as I wrote yesterday, this is the time to take attendance off of the names on the tests. Unlike video notes, students will actually complete the test.

The only real mistake I make today is forgetting to have fifth period place the desks back into groups before they leave. And so I'm stuck moving desks during lunchtime before I can leave.

Speaking of tests, today is the Geometry final exam -- and that test won't be delayed either. This is finals week, and as is my tradition, I post the final on the first day of finals week. Because it's a test day, it's also time for another "traditionalists" post.

(In case you haven't figured it out, today I subbed in my new district, where this isn't finals week. It's only Day 74 in this new district.)

For today's traditionalists' post, I've been thinking about the Twitter account @CCSSIMath. I've mentioned this tweeter before -- the tweeter appears to be a traditionalist who is highly critical of the Common Core Standards for not being rigorous enough.

https://twitter.com/CCSSIMath

Yesterday, CCSSIMath tweeted two Pythagorean Theorem problems -- the first a Common Core problem, the second based on how math "ought to" be taught. Here's the Common Core problem:

A man is trying to zombie-proof his house. He wants to cut a length of wood that will brace a door against a wall. The wall is 4 feet against the door, and he wants the brace to rest 2 feet up the door. About how long should he cut the brace?

But this zombie problem is "zombifying dull," according to CCSSIMath. Instead, the tweeter prefers questions like the following:

A quarter circle is inscribed in a 45-degree sector. Determine the ratio of the radius of the sector to the radius of the inscribed quarter circle.

Let's try solving this problem together. In the original post the points aren't labeled, so we will label them ourselves. Let O be the center of the 45-degree sector, with A and B the endpoints of the arc that forms the boundary of the 45-degree sector. Then let P (on OA) be the center of the quarter circle, with C (between O and P) and D (on arc AB) the endpoints of the quarter circle boundary. Finally, E is the point on OB that is tangent to arc CD.

Since we're asking for the ratio of the radii, we can let the smaller radius be 1, so that our goal is to find the larger radius. Now this is the tricky part -- any point on circle P is now 1 unit away from P, including C, D, and E. So which one do we need for this problem?

As it turns out, PE = 1 is the first radius we need. Now as a tangent is perpendicular to the radius at the point of tangency, OE is perpendicular to EP, so OPE is a right triangle. Also, Angle POE is 45, since it's the central angle of the 45-degree sector. Thus Triangle OPE is a 45-45-90 triangle with leg PE = 1, and so its hypotenuse is OP = sqrt(2).

Now we use another radius of circle P -- PD = 1. Now Angle OPD is 90, since it's the central angle of the 90-degree sector (the quarter circle). So Triangle OPD is a right triangle with two of its legs known -- PD = 1 and OP = sqrt(2). Thus by the Pythagorean Theorem, OD = sqrt(3) -- and OD is indeed a radius of circle O. Therefore the desired ratio is sqrt(3) (or sqrt(3) : 1, to make it a ratio).

According to CCSSIMath, this should be considered an "eighth grade problem." But notice that we needed the Radius-Tangent Theorem (Lesson 13-5) to form one of the right triangles needed to use the Pythagorean Theorem. But the Radius-Tangent Theorem isn't normally proved used in eighth grade math (whether before or after Common Core). But even if we assume that eighth graders have learned the Radius-Tangent Theorem, I wonder whether they can really solve this problem.

This is the sort of problem that CCSSIMath wants to see more of in math classes -- and this tweeter criticizes Common Core for not encouraging this type of problem more. This traditionalist doesn't like Pythagorean Theorem problems where students are given two sides of a right triangle and asked to find the third -- that's "zombifying dull." This traditionalist insists that students work hard and think a little to find the right triangles and their side lengths.

It's still December -- the month of the Putnam exam. As a former participant, I know how why the test takes six hours to answer 12 problems -- many problems require a half-hour, or even an hour, to find the sudden insight, the clever move, the "Aha" moment, that leads to a solution.

I admit that it took me over an hour to solve this "mini-Putnam" problem. At first I did find the value OP = sqrt(2), but then I kept thinking that OE = 1 was significant. In fact, I thought about the point F on ray OA such that Triangle OPF is another 45-45-90 triangle, with hypotenuse OF = 2. By the drawing, it's easy to see that A lies between O and F, and so we know that the desired radius must be less than 2. But then I kept trying to go down this blind alley, believing incorrectly that I was close to finding the desired radius.

This problem frustrated me -- so how much more than would an eighth grader be frustrated? Yet this is want CCSSIMath wants to see more of in eighth grade math classes.

Indeed, I wonder what an actual math class would look like under CCSSIMath's vision. For example, would eighth graders receive ten mini-Putnam problems for homework similar to this problem? If each takes an hour to solve (as last night's problem did for me), then that's ten hours of HW. Hmm, perhaps only one problem should be assigned. But I suspect eighth graders are more likely to guess and make up an answer than actually spend an hour coming up with the clever trick.

Then maybe the mini-Putnam problems should be assigned in class instead. That is, the class begins with the teacher asking this sort of question, and the students have all period to solve it. Would the students work quietly and independently, or should they discuss their answers in groups?

How often would these questions be assigned? At some point, the teacher would actually need to teach the Pythagorean Theorem and Radius-Tangent Theorem needed to solve the problem. (We're dealing with a traditionalist, CCSSIMath, after all -- and traditionalists like direct instruction.)

And what would the tests look like? This is finals week, after all. Should these mini-Putnam problems appear on the final? If so, how many such problems should appear? On the actual Putnam, most contestants score zero, because they can't come up with the clever move needed to reach a solution to any problem. Do we want really most students to score zero on the final?

CCSSIMath is definitely a traditionalist whose tweets I'll be looking out for more often -- but once again, there's only so much one can say in a tweet. At least there aren't any mini-Putnam problems on the final that I'm posting today.

But now I must post the first semester final exam. Here are the answers to the questions on the final:

BBBCA AADAB BBADA CADBC ABBAA BDDAC DDBCC ADBBA BDABC ADDCA









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