Quadrature of the Lune

My sources:  A great introduction to the concept of quadrature can be found on Wikipedia HERE. 

And of course, Leonardo da Vinci was fascinated with "squarable figures."  A fun couple of puzzles that you can think about and even solve are HERE. 

But because I grew up pre-internet, I go to books.  Leonardo's Dessert: No Pi, written by Herbert Wills, III and published by the National Council of Teachers of Mathematics in 1985, was my first venture into quadrature.  Also, Will Dunham wrote a book with a first chapter that discusses Hippocrates' Quadrature of the Lune: Journey Through Genius, published in 1990.


Both Hippocrates and Leonardo da Vinci were interested in squaring "lunes" or regions defined by two intersecting arcs.  I believe each of them viewed this task as a fun pastime -- puzzles to be created and solved for great entertainment.  In class, saw and rectified the following images:

The Pendulum, rectified:

 The Axe, rectified:

If you remember, the Axe appeared in the 2001 SAT test.  Feel free to share this blog with your parents; I used this area problem on Parents' Night last fall.

Let's start by considering The Lune of Hippocrates.

To construct the lune, start with a circle with two radii drawn at right angles and the chord connecting the points on the circle met by the two radii, as shown. Find the area of sector AOB; it's one quarter of the whole circle with radius r.   (A = pi r^2/4)

Finding the midpoint of segment AB gives us the center of the circle with diameter of segment AB, shown as the red circle below. Find the area of  half of the red circle.  (A = pi r^2/4) Notice it's the same as the quarter of the larger circle above.

The Lune of Hippocrates is in the picture above; I've shaded it in the image below:

You could find the area of this shaded lune by taking half of the red circle and subtracting the section of the large circle defined by chord AB.  (Remember that a "section" is a region defined by an arc of a circle and its chord.)

Notice that if you took the quarter of the larger circle (same area as half the red circle) and subtract the same section defined by chord AB, you'd get triangle ABO.

Therefore, triangle ABO has the same area as the blue shaded lune.

Consider for a moment what's happened. You've found two regions that are quite unlikely candidates for having the same area as having the same area. The lune shaded above has the same area as triangle ABO. 

For your blog, find another puzzle to rectify, and rectify it.  You can use the links I've provided above for my sources.  (The Lune of Alhazen might be a good choice in Wikipedia.) You can also look into the history of rectification of shapes or perhaps how that applies to ideas in calculus. Google away, find something to teach me.  Have fun with this one!  If you're a puzzle person, play with a puzzle. If you're a history buff, then look into how this idea of rectification really furthered some mathematical thought.  Or if you are an artist, see if you can find how Leonardo or anyone else used these ideas in art.

Here's a sample of one puzzle with its solution, for a demonstration. It's a challenging solution, so don't feel as though you need to do one this challenging.  Consider a "Motif" that Leonardo used, and the corresponding region that I've shaded:

And here is how you might rectify that shaded region (at least, you'd cut and rotate until you get a pendulum, which you know to be rectifiable).