Saturday, May 16, 2009

The "POWER" of Circles Part I - An Open-Ended Geometry Challenge and CONTEST Update

Well, registration for MathNotation's 2nd Online Contest has now closed. Not as many participants this time but we do have schools representing several states and one high school from Japan! The questions have been emailed to advisors but results will not be available for a couple of weeks. I also plan on publishing at least one of the contest problems in June. I'm still in the planning stages for running 2-3 of these contests for the 2009-10 school year. Stay tuned...

If you're interested in signing up for upcoming contests, just drop me an email at "dmarain at gmail dot com" and I will put you in my database.

For those of you who haven't been reading about these contests it's the team approach, open-endedness and multi-part nature of some of the problems which separates these contests from most others out there. In other words, these questions reflect the types of investigations I've been publishing since 2007.

Also, after I write 5-10 of these contests, I plan to publish these in a book with detailed solutions and comments. As my wife would say, "I'll believe that when I see it!"
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Speaking of investigations, here is Part I of an open-ended geometry problem that starts out relatively simply but eventually will lead to deeper results. This investigation will review basic circle concepts involving tangents and secants but will connect to the more advanced ideas of power of a point and the inverse of a point in a circle in later posts.

In the diagram, O is the center of the circle PT is a tangent segment, segment AB is a diameter.

(a) If the radius is 6 and PA = 4, show that the length of tangent segment PT is 8.

Note that (PA)(PB) = (4)(16) = 64 and, from part (a), (PT)2 also equals 64. This is a special case of the secant-tangent power theorem you may recall from geometry. Your job in the next part is to demonstrate a particular case of this theorem this using the above diagram.

(b) If the radius of the circle is r and PA = x, show that
(PA)(PB) = (PT)2.

Note: Using the secant-tangent power theorem here trivializes this problem. The idea is to demonstrate the result without using that theorem, in effect, proving a special case of this rule!

Click on Read More for further comments and a hint for part (b)...

(i) Neither of the above parts was intended to be highly challenging. Part (a) is definitely an SAT-type question. A review of geometry never hurts!

(ii) Here's a hint to get started on part (b):
From Pythagorean we know that (PT)2 = (PO)2 - r2. Factor this expression...

(iii) Note that my approach as always is to introduce or develop a theorem or concept such as the secant-tangent power theorem or "power of a point" by looking at particular or special cases (the secant in the diagram contains a diameter) and starting with numerical values rather than variables. This sequence (numerical values, particular case, generalization) forms the basis of the investigations I've been writing for this blog, but, more importantly the underlying foundation for the lessons I planned when I taught. I believed then and now that this approach may be time-consuming (both in planning and implementation) but the payoff is deeper conceptual understanding. Of course I needed to modify the presentation according to the backgrounds and ability levels of my students but I never assumed only my honors and AP classes were up to the challenge.

(iv) In Part II which I will post in a few days, we will discuss the "power of a point" and add a second circle, in which segment PT will be a radius. This will produce a another point, P', the inverse of point P.

Rohan Sharma said...
This comment has been removed by the author.
Anonymous said...

If the radius of the circle is r and PA = x, show that
(PA)(PB) = (PT)2.
Isn't all that r and x business superfluous?

Dave Marain said...

You're right Anon--
PT^2 = PO^2 - r^2 = (PO+r)(PO-r) = (PB)(PA).
You don't need to name PA by using x at all. However, naming the radius r makes sense to me here. In Part II of this investigation, we will define (PT)^2 as the power of point P and show there are alternative formulations using the discussion in Part I. Then we will locate a 2nd point P', called the inverse of point P, such that (PO)(P'O) = r^2. That's why I needed r. If P is "on" the circle it's apparent where P' would be!