Playing around with Euclid and constructions

I was playing with Euclid’s Elements perhaps two months back when I had some spare time (or rather when I was bored with whatever it is I was supposed to do). I had never actually read it before and I guess there are two reasons for it. Firstly there aren’t any modern translation of it into Swedish and the older ones read like and equivalent of Shakespearean English with a hard to read way of spelling so to come into contact with it as a kid is not impossible but nevertheless hard. Also in school geometry isn’t taught in the spirit of Euclid with more of an emphasis on using geometry formulas to teach algebra and once trigonometry enters the picture Euclid presentation seems somewhat irrelevant.

Having read the first book and skimmed through some parts which I was interested I’m not yet in the camp that yearns for the return of ‘axiomatic geometry’ in schools since even though I find the presentation appealing I still agree more with the camp that thinks that education is supposed to be organized to developing useful skills and — yeah I’m not convinced that this is the case. But it is interesting and I can perhaps appreciate the structure more now that I have a better grasp of how mathematics works in general.

Still, what I’ve appreciated the most looking into this area are the construction (the arguably least relevant apsect of it). I have a a special place in my heart for ‘proof by construction’ and of course the Elements is rife with them. Even if you discard the proof aspect of geometry just sitting and playing with the (virtual) straightedge and compass turns out to be ridiculous amount of fun. To start with the question of ‘Can I do it?’ and then you play around with the tools to see what you end up with. Just two tools end up yielding absurd levels of complexity and compared with the 5+ operations of algebra it’s arguably simple.

This phase of trying out straightedge construction has also coincided with me wishing to learn how to draw graphs in TikZ (QTikZ) so I get to practice it through outlining construction schemes.

For example one of the first questions I asked was how to find the circle which intersects three arbitrary non-colinear points in the plane with the compass and straightedge. The reason that was the first question I asked was really because it’s the first computer lab you do at KTH (my school) utilizing linear algebra. Then it has a follow up question of taking an arbitrary number of points with and you’re supposed to find an approximate circle with the least square method but let’s ignore that.

What I’ve found so lovely working with these tools have simply been that the constructions almost always seem to practically write themselves once I got a hang of the rules of the game. At least to me.

In this case we of course start with developing the basic operation if finding a perpendicular to/bisecting a line which follows from the first proposition in the book.

Constructperpendicular

Bisecting a line or drawing perpendicular

Now this is of course obvious to everyone (especially Americans who I’m told are actually told who Euclid was/might have been in school) but these things aren’t in the Swedish curriculum anymore so to me it’s been like discovering something new and beautiful. You get this almost childish joy from seeing it when you look at it as puzzles rather than formal mathematics.

Now for finding the circle which intersects three points I really started from a simpler question of how one could recover the center of a circle in case it had been lost. For me the fundamental property of a circle is not really its definition but that it is a perfectly symmetric shape and so in cutting a circle (disc) in half I should always end up cutting it along a diagonal. From this principle the following procedure using a cord for finding the circles center is to me very natural

Image

Finding the center of a circle

A proof that it works can be made but its intuitively obvious that it works even without one.

From this, the idea that a perpendicular to a chord passes through the center of the circle, it is immediately obvious that the perpendicular to two different chords should intersect at the center. This simple idea is all that you need to at least guess that the following procedure using two virtual chords should construct a circle.

Image

Finding a circle which intersects three non-colinear points

Of course at heart, or in the way I at least thought about it while doing it, this was really not a proof of that you can draw a circle through any three points but rather that you can reconstruct a circle from three points that were taken to lie on an ‘existing’ circle. But as I said these intuitive constructions seem to always come with the lines necessary for forming a formal proof and by just drawing 3 extra lines I can complete it to a formal existence proof with a simple argument.

Image

Verifying that circle passes through all three points

The uniqueness part is also simple to see as the two perpendiculars can only meet at one point in our geometry. (All this is probably in Euclid but I’ve been more into solving stuff for myself than reading through it)

I’ve done this with a tonne of different constructions now and the overall feeling I’ve gotten is that this really should be quite easy to turn into a puzzle game. Maybe I’m wrong but so long as the constructions are put in a certain order (the way I discovered them in this case) it could be a pretty entertaining game. Maybe you need to be able to see the nature of the corresponding proofs for it to work but I wouldn’t think so. I think it should work just fine.

Let me just round off by outlining the algebraic solution to the question of the existence, uniqueness, and construction of a circle from three points. Let (a_1,a_2), (b_1, b_2), (c_1, c_2) be the coordinates of the three points. Then any center (x,y) of a circle with radius R intersecting all three points must satisfy the three equations

(x - a_1)^2 + (y - a_2)^2 = R^2

(x - b_1)^2 + (y - b_2)^2 = R^2

(x - c_1)^2 + (y - c_2)^2 = R^2

which may  be cast into linear form by subtracting the second from the first and the third from the second to give us

2(b_1 - a_1)x + 2(b_2 - a_2)y = b_1^2 + b_2^2 - a_1^2 - a_2^2

2(c_1 - b_1)x + 2(c_2 - b_2)y = c_1^2 + c_2^2 - b_1^2 - b_2^2

discarding the third. Then we have linear system which is solvable with a unique solution if the points are not colinear as can be seen from either the determinant of the fact that the two columns in the corresponding matrices are linearly independent if that is the case. The radius can be recovered from the distance between (x,y) and one of the other points.

This is arguably the simplest solution since all issues of construction, existence and uniqueness are put into a single generalization procedure but you also need much more advanced machinery to do it. The other one takes next to none.

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