Circular Euclidian Geometry: Overlooked Creativity

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Solutions to the Paradox Problems:

1. There is still more space to fill. Assign all individuals currently residing in the hotel to even hotel room numbers. Assign all individuals who have arrived at the hotel to odd hotel room numbers.

2. There would be no cards left.

3. The wizard would see all his coins return to him, and the mermaid would have no coins. The wizard and the mermaid would have equal amounts of coins. The wizard would lose all his coins, and the mermaid would have all the coins.

If you would like to see how I arrived at these conclusions, just ask.

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“Do not disturb my circles!” — Archimedes

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I was originally planning to discuss infinity in its entirety today, but in wake of Hurricane Sandy, I felt inspired to do a short activity on circular Euclidian geometry. (After all, hurricanes are roughly circular in shape.)

We are all familiar with A = πr², the formula to find the area of a disk bounded by a circle of radius r. (Notice I said area of a disk and not area of a circle. If you ever hear your friends say “area of a circle,” correct them immediately. A circle is a set of points equidistant to a common point. You cannot find an area of a set of points equidistance to a common point!) Despite being a simple formula, most high school students still have no idea how this amazing conjecture was discovered. Today, we will “prove” the theorem to be correct, and hopefully, you will come to see and appreciate how much creativity can be compacted inside a simple formula like A = πr². (You will need a piece of paper!)

Draw a circle on your paper. Cut that circle into four equal sectors (i.e. four equal slices of pizza). Remove two of these sectors from the circle and arrange them such that both sectors point upwards. Take the remaining two sectors from the circle and arrange them such that both sectors point downwards. Place these two sectors above the sectors pointing upwards. Naturally, you can see that the two pairs of sectors will fit nicely into each other to form a quasi-parallelogram.

Now, repeat this process by creating another circle and cutting it into eight equal sectors. Remove four of these sectors from the circle and arrange them such that the sectors point upwards. Take the remaining four sectors from the circle and arrange them such that the sectors point downwards. Place these four sectors above the sectors pointing upwards. Again, notice that these two groups of sectors will fit nicely into a quasi-parallelogram. The fit, however, is more refined this time than it was last time.

For clarity purposes, let us repeat the process one last time. Create another circle and cut it into sixteen equal sectors. Remove eight of these sectors from the circle and arrange them such that the sectors point upwards. Take the remaining eight sectors from the circle and arrange them such that sectors point downwards. Place these eight sectors above the sectors pointing upwards. As before, the pieces fit together, and we see the same quasi-parallelogram that we saw in the last iteration. This time, however, notice that the quasi-parallelogram appears more like a rectangle. This is great news because we know the area of a rectangle: A = bh, where b is the base of the rectangle and h is the height of the rectangle.

So now, imagine that we extract infinitely many sectors from a circle. Half of those sectors will point upwards. Half of those sectors will point downwards. If we fit these two groups together in the manner we have been doing, theoretically, we should get a perfect rectangle. Therefore, all we need to know now is the base and height of that rectangle.

The base of the rectangle is a bit hard to see at first, but take notice that it is comprised of tiny arclengths from the circle. These arclengths add together to give half the circumference (“perimeter”) of the circle with which you started! Because we know that the circumference of a circle is given by the equation c = 2πr, where r is the radius of the circle, the base of the rectangle must be half that value, or b = πr. (If you cannot see this, keep looking at your diagram! It will click!)

The height of the rectangle is easy to see because it is given by the lengths of the edges of the sectors from the circle with which you started. Notice that the edges of these sectors have a length equal to the radius of your initial circle. Therefore, h = r.

Substituting both of these values into the formula A = bh, we get A = (πr)(r), which simplifies to A = πr².

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Mathematician’s Side Note

Note: I am assuming you are familiar with the following concepts.

Another way to find the area of a disk with radius R, called the “Onion Peeling” method, uses basic techniques from calculus. Consider a disk to be nothing more than a series of concentric rings. Each of these rings has a small area associate with it, and if you were to cut one of these rings and stretch it out horizontally, you would see that the ring would actually become a small rectangle.  The area of this rectangle could be given by the product of the rectangle’s height and base. The base of the rectangle would be equal to the circumference of the ring. The height of the rectangle would be given by an infinitely small change in the disk’s radius. In other words, the small area of one ring would be equal to the product of the ring’s circumference and the infinitely small change in the disk’s radius, symbolically (dA) = (2πr)(dr). We now wish to add the areas of all the rings from radius zero to radius R, the disk’s radius. Symbolically, we would write Int(2πr, r, 0, R). Evaluating this integral leaves the result A = πR², which is what we sought to prove.

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