Geometry, Desktop Environments.

So, here we are at this very important juncture in life. Probably three months from now, I shall be clicking “Send” on each of my college applications. But first an update on what I have been doing. I am trying to write clean code and create a backup script. It seems to be coming on nicely. I have to just include the command line options and it should be in working condition. Apart from that I have switched to xfce which seems to be a terrific alternative to KDE (ahem for my museum piece). I don’t know why but I happen to dislike KDE. It is such a resource intensive envrionment and comes with unnecessary bells and whistles like bouncing icons and whatnot. XFCE is ok.

I still love fluxbox. By the way I came across a wonderful book called “Geometrical constructions with compasses only” by A. Kostovskii ( I don’t know why the author put two “i”s in there).

The Mascheroni theory states that, “All construction problems solvable by a ruler and a pair of compasses can be solved by a pair of compasses alone. ”

I had a look at some sample problems in the book. Most were of the variety that I had done already such as drawing line segments n times the length of the given line segment and so on. I skimmed through and got to the part that I hadn’t come across before.

Inverse. Let us say we have a circle with centre O and radius “r”. If we fix a point P somewhere in the plane of the circle, then point P’ is the inverse of point P if |OP|.|OP’| = r ^ 2. The circle in question is called the circle of inversion.

I then had a look at some of the problems that came at the end of the section on the inverse of a point. One question involved proving a theorem that went:

“The inverse of a line with respect to a circle is a circle.” Let us see how this can be solved. I felt that this was very simple. I fixed a point P on the line, dropped a perpendicular from the centre of the circle on the line and marked the point of intersection as Q. I then plotted the inverse points of those two points. Considering that Q is the foot of the perpendicular from the centre, we can safely assume that the angle OP’Q’ is equal to angle OQP = 90 degrees as OP’Q’ and OQP are similar triangles. As P is moved along the line, P’ moves along the circle as the angle OP’Q’ will always be 90 degrees.

The next question involved plotting the inverse of a point P with respect to a circle with centre O using only a compass. I used the idea of similarity as quoted in the previous paragraph. If P’ is the inverse of P, OP/OQ (Q being a point on the circle) = OQ/OP’. Finished. OP’ x OP = OQ ^2. So if we draw a circle with centre P such that it cuts the circle at points D1 and D2, With D1 as centre and OD1 as the radius, draw an arc and cut this arc with another one obtained by carrying out the above procedure with point D2. Tada, inverse obtained.

I then had a look at some sample problems solved by Mascheroni himself !!

One feature I have observed about the constructions done by Mascheroni are very elegant. For example have a look at this question:

“Plot the centre of a given circle.”

I rattled my brains and came up with the age old solution of getting two normals to intersect. I simply could not avoid using the ruler. One would of course attribute this to the fact that my experience with the inverse of a point or a curve is limited but I feel that if ideas don’t strike when the material learned is fresh in the mind, when will they strike?  I hate looking the solutions to problems but I was forced to look at the solution to this one. Here goes:

Take a point on the circle. We’ll call this point “O”. With O as the centre and taking any arbitrary radius “r”, draw an arc cutting the given circle at A and B. With OA as radius, draw an arc and cut this arc with another one drawn with B as centre and OB as radius. We should be getting point O’. If we draw a circle with centre O’ and radius OO’ such that it cuts the first drawn circle at D and D’, the points of intersection of arcs with centres D and D’ and radii OD and OD’ respectively give us point O” which is the centre of the given circle.

The proof seemed fairly simple as soon as I read the solution. Since A and B were lying on the same circle, A was the inversion of B and vice versa. Hence the circle is the inversion of line AB. So elegant.

I don’t know when I will reach the capability of Mascheroni, V. A. Uspensky and other mathematicians whose works I have been exposed to. I just hope that that day comes soon.