Monday, June 16, 2008

A Geometry Classic - Chord and Tangent Riddle


Don't forget to submit your solution to this month's Mystery Mathematicianagram (ok, so I can't decide on a name yet!). We've received 3 correct solutions thus far and I will announce winners around the 20th.







As we wind down the school year, the problems below may come too late for students taking their final exams in geometry, but you may want to hold onto this classic puzzler for next year. I don't consider these overly challenging but I do feel they demonstrate some important mathematical ideas and problem-solving techniques. Further, encourage students to justify their reasoning since some may make assumptions from the diagram without verification. This will review some nice ideas from circles.

OVERVIEW OF PROBLEMS (see diagram)
For both questions, assume the circles are concentric, segment PQ is a chord in the larger circle and tangent to the smaller.

PART I
If PQ = 10, show the difference between the areas of the 2 circles is 25π.

PART II (the converse)
If the difference between the areas of the circles is 25π, show that the length of PQ must be 10.

Notes
(1) It is important for students to recognize that there are many possible pairs of concentric circles (varying radii) satisfying the hypotheses of these problems, yet the conclusions are unique! Some students will assume a 5-12-13 triangle is formed (not a bad problem-solving strategy), but stress that this is not the only possibility!
Remember, we're not restricting the radii to integer values.

(2) There is a classic math contest strategy for these questions that mathematicians love to employ - the "limiting case." Can you guess what I mean by this phrase?

4 comments:

Florian said...

Nice little problem! I especialy
like the fact that the center of
the circles in not drawn. That
forces students to think :)

Alex said...

Well, the proof is easy enough. Let the centre be O, the radii be R and r, and the point where PQ is tangent to the smaller circle be T.

Then OT = r, OP = R and OTQ = 90 degrees
So PT = sqrt(R squared - r squared) by Pythagoras
Thus PQ = 2*PT = 2*sqrt(difference between areas/pi), and the result follows.

By the limiting case, you mean where r=0... then PQ = 2R and the difference between the areas is simply the area of the large circle.

What I can't come up with is a mathematically satisfying proof that develops from that limiting case. Did you have one in mind?

Dave Marain said...

Florian and Alex--
Thanks for the comment and the solution. I'm really interested in how students at different levels might approach the question. For me, the pedagogy and issues of how youngsters think and develop are of paramount importance.

The term 'limiting case' might be a bit misleading here. If we assume a unique answer to these problems, then the result should be independent of the radii R and r. Therefore, we can use the method of 'particularization' and choose R = 5 and r = 0 (or for that matter radii of 13 and 12 since that also produces a difference of 25pi). The 'limit' issue may arise if one does not accept that r = 0 is a valid choice, in which case, I guess I would use limit theorems to justify it.

Certainly, there's some nice intuition involved when we ask students to describe what PQ and T are approaching as r --> 0. That's not rigorous of course but it seems appropriate for the age group.

hashimi said...

I did it in the following way
let "r" be the radius of bigger circle and "r' " be the radius of smaller circle. now the difference between areas of any two concentric circles is
Area of bigger cirlce - Area of smaller circle
which yields the following formeula
Diff. in Area = Pi * (r^2 - r'^2)------> (1)
Now as PQ is the Chord in bigger circle so joining the center of circle O to the mid point of the chord M will give us PM which is perpendicular to PQ. Now joining the center O to the point P, will give us a right angle triangle. so applying pythagaorus theorem we get
r^2 = r'^2 + (PQ/2)^2
r^2 -r'^2= 25 because it PQ/2 =5
using this value in equation 1 we get
Diff. in Area = Pi*(25)
=25*Pi
Well I did it in this way. probably there may be others.