Another summer diversion from geometry...
The number of variations for tangent circles is endless and this is one of my all-time favorites. Math contests and SATs seem to have a preference for circles inscribed in squares or tangent circle problems and this one is along those lines. However, the real payoff comes from developing recursive thinking leading to an infinite geometric sequence and its sum! Students will be asked to intuitively "guess" the value of this infinite sum and to then verify their conjecture. Proving it requires nothing more than the classic formula for the sum of an infinite geometric series but, at the outset, this problem is eminently suitable for your geometry classes. Don't hesitate to use it in your "regular" classes. Questions that are deemed appropriate only for honors classes are often suitable for most students if the groundwork is laid (background, examples, etc.) and hints are given strategically.
PART I In the diagram above the larger circle has radius 1, the two circles are tangent to each other and to the two perpendicular segments (you can think of the larger circle being inscribed in a square if you wish).
(a) Make a conjecture from the diagram without computing: The ratio of the radius of the smaller circle to the larger is approximately
(A) 0.05 (B) 0.15 (C) 0.25 (D) 0.35 (E) 0.5
Note: This part may be omitted.
(b) Show that the radius of the smaller circle is exactly (√2 - 1)2 = 3 - 2√2
How was your conjecture?
Note: Your decision about giving them the result like this. Obviously if they see part (b) on a worksheet, their estimate in part (a) will be pretty good! My intent was to focus on the method. Of course, feel free to rephrase this.
PART II
Of course we will not stop at 2 circles! Squeeze a third circle into the corner between the 2nd circle and the right angle. Determine its radius by using the result from part (a). [The key here is to think ratios!]
PART III
If we label the radius of the largest circle R1, the radius of the 2nd circle R2, the radius of the 3rd circle R3, etc., we can now define an infinite sequence of these radii.
(a) Find a formula for the nth term of this sequence, n = 1,2,3,..
(b) What is the mathematical terminology for this type of sequence?
(c) Think intuitively here: From the diagram, what should be the "sum" of the original radius R1 = 1 and the diameters of the remaining infinite collection of circles. [Another formulation: As n-->∞, this sum approaches what number?]
(d) Using the formula for the sum of an infinite geometric series, verify your conjecture in (c).
Comments:
- As always, feel free to use this with your students and revise as you see fit. However, pls use the attribution in the Creative Commons License as indicated in the sidebar.
- Finding the radius of the 2nd circle is a challenge by itself and the problem could stop there. The extensions can be assigned as a long-term project or for those wishing to do extra credit. I always liked having additional challenges for the students who were capable of going further, although relating this problem to geometric sequences or series is of importance. Of course, I am well aware of time constraints faced by the instructor.
- Your thoughts...
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9 comments:
That's a good problem. Students would need to use similarity and methods for simplifying quotients involving radicals.
For a source of similar (and more difficult) roblems, consider the arbelos (on MathWorld).
Very tedious, using the quadratic formula and all...unless there's a shortcut? That would be nice.
Getting additional radii of smaller circles is a breeze after getting the first smaller one.
An interesting aspect to consider (that i'm not quite sure of): The other root you get from using the quadratic formula is 3 + 2 root 2, which is around 5.828427125. Could that have any significance? Perhaps in another dimension considering another axis? I'm not sure.
You don't have to use the quadratic formula on these problems do you? i mean you just have to think about it logically. the quadratic equation only complicates things.
Hmm i guess I'm not up to that level of thinking. The quadratic formula was the first thing that popped into my mind; I was raised with it.
Yzw731, look at how the line y = x intersects the large circle. It strikes the circle at (1±√2, 1±√2). Now, it strikes the small circle at (t, t) and (1-√2, 1-√2). For some t. Use similarity to find t.
Sorry, yzw. This is what I meant. Call the large circle C and the small one C'. The line y = x intersects the circle C first in point A (1-√2, 1-√2) and then in point B (1+√2, 1+√2). It passes through the center O (1, 1).
Similarly, the line meets C' in A' and B' = A. It passes through its center, O'. But, the configuration of the origin, axes, and C is similar to the configuration of origin, axes, and C'. So, corresponding lengths are proportional. The radius of C' is to the radius of C as the height of B' is to the height of B. So,
r_{C'} : 1 :: 1-√2 : 1+√2.
Solve for the radius: r' = (1-√2)/(1+√2). Clear the radical from the denominator, and get 3 -2√2.
Thanks for you ideas!
Thanks, Eric, for adding the clarification.
A few thoughts.
(1) When I computed the pts of intersection of y=x with the circle of radius 1, I obtained (1±√2/2,1±√2/2). This can also be seen from drawing the 45-45-90 triangle up and to the right from the center (1,1). The hypotenuse is the radius 1, so the legs would have lengths 1/√2, etc. However this does not affect the ratio calculation for the 2nd and subsequent radii! Showing the equivalence is a nice exercise on radical simplification.
(2) I think your similarity explanation is crucial here and important for students to see. Recognizing that a sequence of these similarity relationships is equivalent to a geometric sequence is something that should be stressed more.
(3) Another approach which is less based on ratios and proportions is to consider the following calculation which will be hard to follow without a diagram:
Let r be the 2nd radius and construct a square from the center of the smaller circle by drawing horizontal and vertical segments to the "axes."
Then we can express the diagonal of the square from (1,1) in 2 ways:
r√2 + r + 1 = √2.
Solving for r produces the desired result, r = (√2 - 1)^2. Repeating this procedure using (√2 - 1)^2, produces the 3rd radius and so on. This leads to a geometric sequence with common ratio (√2 - 1)^2, etc.
It's amazing how one can make a mistake and not see it. Of course 1-√2 couldn't appear in the problem. I'd rewrite my comment, but I'd probably make similar mistakes doing so.
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