Tuesday, October 30, 2007

The Power of Visualization: Another View of Inscribed vs. Circumscribed


A student, Melissa, in my SAT class, shared her way of thinking about a well-known problem that appeared on this blog awhile back as part of a larger investigation. You may recall the post about the ratio of the areas of circumscribed and inscribed figures.

I told her I would celebrate her method on my blog, which would definitely be viewed by less than or equal to a million people a day! One of my students asked me why I don't simply write a book about all these ideas and I replied, "I'm thinking about it." He asked what the title would be and I replied, "What I've Learned From My Students."

Here's the problem/investigation/challenge/activity///// for you or your geometry students:

Consider the diagram above. Assume that it depicts two squares. The smaller inscribed square is formed by joining the midpoints of the larger square.


(1) Explain why the area of the inscribed square is one-half of the area of the larger.
Easy so far...

(2) Consider the circle circumscribed about the smaller square, i.e., it passes through its 4 vertices. Explain why this circle is inscribed in the larger square. This requires that you show that the 4 sides of the larger square are tangent to the circle. [By the way, most students would assume this is obvious from the diagram, but ...]

Not impressed? Deja vu all over again as Yogi would say? This is how Melissa demonstrated that, for a given circle, the area of the circumscribed square is twice the area of the inscribed square. Is this the method you have seen or used yourself? Consider that she used no variables, didn't plug in particular values for the dimensions, etc. She drew the diagram and basically said that the diagram proves itself! A proof without words, so to speak (that might even make Sidney Kung proud). I congratulated this young lady and the class applauded. Interesting how students immediately recognize someone's brilliance...

14 comments:

Totally_clueless said...

Impressive! The force is with her!!!

Dave Marain said...

I'm glad you appreciate what she did. I demonstrated the traditional way of showing it:
I let 2r denote the diameter of the circle and explained that this was both the diagonal of the small square and the side of the big square.

She timidly raised her hand and said, "I think there's an easier way," to which I replied, "Show us!"

Jackie said...

I love it!

Totally_clueless said...

However, the pedant in me refuses to go away. For her method to work, you also have to show that all in-squares and circum-squares (and there are millions of them) have the same areas.

TC

Dave Marain said...

tc--
I think, I'm missing your point. For a given circle, there is a unique 'circum-square' and a unique inscribed square. Conversely for a given square, there is a unique inscribed circle, and, corresponding to this circle, there is a unique inscribed square. All such combinations of a circle and its 2 squares, are, of course, similar to each other, so the ratio of the areas is invariant.

Pls clarify for me...

Totally_clueless said...

Given a circle, are the circum-square and the in-square unique? (they are only unique upto the dimension of the square, but certainly not in the orientation). The proof presented is true for one combination of in-square and circumsquare, where the vertices of the in-square are the mid-points of the sides of the circumsquare. To show that this is true for all in-squares and circum-squares, don't you have to show the uniqueness of these in dimension?

It is a pedantic duscussion (some may use a less generous anatomical description :-)), but I am wondering if the proof as presented is mathematically complete.

TC

Totally_clueless said...

And if you consider inscribed rectangles, they are certainly not unique.

TC

Dave Marain said...

tc--
I would NEVER characterize any of your contributions as pedantic! What you're questioning is the mathematical validity of the argument. A proof by diagram is not mathematically precise and, now that you've clarified your point, I completely agree that more discussion is needed here.

Another approach might be to start with a given square. There is then a unique inscribed circle. Now it gets sticky. There are infinitely many inscribed squares as you indicated. Consider the rigid 'frame' formed by the circle and any one of these inscribed squares. Rotate this 'circle-square' rigid frame about the center until one of the vertices of the inner square coincides with one of the points of tangency of the circle with the outer square. We would need to prove that this orientation is the one Melissa came up with but this requires more argument. This approach does take care of your orientation concern however.

Another approach is to connect the 4 points of tangency and prove that the resulting figure is a square. It's not too difficult to show that the segments connecting opposite points of tangency are diameters and that these diameters are perpendicular. This would then show the quadrilateral connecting the points of tangency is in fact an inscribed square.

I need to think this through, which at this hour of the day is futile!
However, the main point here is that your concern was well-founded. None of this however detracts from Melissa's ingenious approach.

Totally_Clueless said...

I think the straightforward approach is to show that any two in-squares have the same area. Or, take any in-square and show its area depends only on the radius.

Then, we can apply Melissa's insightful argument to the particular in-square, and all is well.

Cheers,
TC

Dave Marain said...

tc--
I definitely overreacted to your concern! Demonstrating that the formula for the area of the inscribed square is 2r^2 is straightforward. I thought you had a much weightier problem with Melissa's argument! Oh well.. Actually, demonstrating that the quadrilateral formed by connecting the 4 points of tangency of the circumscribed square is an inscribed square is not completely trivial and is probably a good exercise for geometry students. One could show that its diagonals are congruent and perpendicular bisectors of each other, but the perpendicularity takes some thinking. Drawing radii to each point of tangency seems to do it but there's a bit more required I think to show that connecting opposite points of tangency form perpendicular diameters. Students don't use those theorems on perpendiculars and uniqueness postulates for perpendiculars as often these days (e.g., through a point not on a line there exists one and only one perpendicular, etc.).

Melissa said...

HI! It's Melissa. (I finally found your blog!) Thank you, Mr. Marain, for giving me my moment of fame.

I would just like to say that my approach could possibly (I'm not sure, as I am still in grade school) be proven if thought of on a coordinate plane.

If a square and circle are rotated within a fixed larger square about a certain point (i.e. the center of the circle and both squares), the square and the circle have not changed, and of course, their radii have not changed length, just position, conveniently to one that can be better measured. My math teachers have told me that translated/rotated/reflected but not dialated figures are congruent, meaning that all of their dimensions are equal. (That has to be a theorem in some book...) Therefore, by rotating the square and the circle, a part of the figure was rotated, and the larger square and the center point have not changed, leaving me with a segment length that has not changed, but is easier to measure.

Dave Marain said...

Glad you found it, Melissa! You deserved that recognition. Your argument is highly mature and makes sense to me.

You are an exceptional student who, in my opinion has unlimited potential in math. If you're not already, you should be participating in math contests and regularly visiting the Art of Problem Solving website, the best source on the web for challenging problems and advice for young mathematicians. Your insights are wonderful and you challenge me to be my best!
Mr. M

Jackie said...

How incredibly cool is that - she went looking for your blog!

Melissa said...

Art of Problem Solving is so cool!!! The little animation on the side is just awesome. It reminds me of Mensa stuff.