Tuesday, January 16, 2007

Problems for 1-17-07


I may have comments about yesterday's problems later. mrc did a great job - read his comments, answers and solutions.

Today's problems (shown above) involve more geometry. They will appear as images (including the text).
These problems review 30-60 and 45-45-90 triangles, radicals, slope and/or similar triangles.

13 comments:

Jonathan said...

I needed an auxiliary segment.

Jonathan said...

The answer to #2 surprised me a bit. I'll let others get to the answer, but I am anxious to discuss the different methods that may be used here.

mrc said...

I used the properties of the special right triangles on #1, and similar triangles (proved by AA) on #2.

I agree, surprising answer for the second!

Dave Marain said...

I'm really glad you enjoyed #2. Of course, many students expect the answer to be 5, assuming an arithmetic sequence. But ratios of similar triangles lead naturally to geometric sequences. Here's something to provoke your students: Have them continue the sequence of squares to the left, producing an infinite sequence of squares. This is worthwhile long before a formal introduction to limits!

Also, I've noticed that many students fail to make the connection between slope and similar triangles. Talk about fundamental concepts!

By the way, I need some advice here. I've actually authored all of these problems and am happy to share them with everyone. How do I convey they are to be used for personal pleasure or for instructional purposes but not for commercial reproduction? Did I just write my own disclaimer!

Mrs. Temple said...

I really like #1... I think I'll use it in my Geometry class :) I wonder how many of them will think about dropping a segment down from E perpendicular to DC so they can have all special triangles :) I'd bet many of them try to work with triangle DEC and get stuck....

mrc said...

The best way to express the conditions on these original problems is with a Creative Commons license: http://creativecommons.org/license/

Darmok said...

Interesting...I liked #2 as well.

Jonathan said...

mrc,

why would Dave need to license these? I have never thought about anything but getting my problems into other teachers' hands. Is there a reason I should be considering this?

Jonathan said...

Did anyone else consider using coordinates for #2? I used similar triangles, but placing the origin at the lower left corner (or better, the upper left corner) of the first square works almost as well.

mrc said...

Two reasons for using the Creative Commons licenses come to mind:

1. He wants these things not to be used commercially. The CC licenses express this in legal language that will hold up in court if necessary.

2. Putting a license on it (especially one that's not too restrictive) makes it easier for people who want to use these problems in a non-commercial context. They can clearly see the terms of the permission being given.

I think it's a good idea for everyone to put a CC license on their blog. (And, now that I mention it, I realize I need to do that!) Not only to protect yourself if you feel like you need it, but also to encourage sharing. CC is all about sharing and making public and clear exactly how you want to share your work.

Dave Marain said...

thank you all for your kind words for these questions!
also, thanks, mrc for the advice -- like jonathan i enjoy sharing the problems i've developed and those who wish to copy these and use them for educational purposes probably should be given the parameters of doing so. I'll take a look at that license and then decide. I don't want anything to discourage viewers.
In the past I have freely shared these with my colleagues but now the number of people who might see these each day is like a geometric sequence whose common ratio is 4/3 or more!!

By the way, drawing an auxiliary segment for #1 is a great idea since it produces a square with a diagonal of 4, so its area is 8, andc then work with the 30-60-90. However, another possibility, without drawing the segment, is to determine the height of the rectangle to be 2rad2 as you all did, then move this height over to the 30-60-90 triangle on the far right and compute the top segment to be 2rad6. Adding this to 2rad2 gives the base of the entire rectangle, etc. This is certainly not easier than splitting it up into pieces.

Dave Marain said...

jonathan,
originally i had this on a coordinate plane, then i decided to see how students would approach it if I left it without a reference system. A couple of them solved it using coordinates but most never even considered it! It takes some nice thinking to decide where to place the origin (very instructive to show them how arbitrary this can be) and how to determine the coordinates but that's the whole point! I think a few were surprised when I said that the value of the slope is very much dependent on where you place your frame of reference! I asked them if the slope of the line could be negative -- that blew their minds!

Dave Marain said...

by the way, folks, when is the best time, in your opinion, for me to post these questions? late at night est? early afternoon of the day before? time is relative when you're connecting to the planet but i'm curious...
If I post too early and the comments with solutions appear rapidly, the interest seems to wane...
Then again I feel it's just a small group of you who are actually trying these!