UTM Ideals Varieties and Algorithm - Chapter 1 Section 3 Exercise 6

Problem:

Solution:

Part a is the most challenging part of the problem, see the drawing below:

The radius of the circle is 1. Therefore, $ |BE| = 2 $, the diameter.

By geometry, we know $ \measuredangle BDE = 90^{\circ} $ because it is the angle subtended by a semicircle, so we know $ \triangle BEC \sim \triangle EDC \sim \triangle EFD $, our goal is to express $ |DF| $ and $ |EF| $ in terms of $ |CE| $.

Once we have the triangle similarity relation, the rest is really just algebra, we know

$ \frac{2}{|CE|} = \frac{|BE|}{|CE|} = \frac{|DE|}{|CD|} $ by $ \triangle BEC \sim \triangle EDC $, and so $ |CD| = \frac{|DE||CE|}{2} $

We also have $ |DE|^2 + |CD|^2 = |CE|^2 $ by Pythagorean theorem, if we substitute the previous result, then we get $ |DE|^2 + (\frac{|DE||CE|}{2})^2 = |CE|^2 $, simplifying, we get $ 4|DE|^2 + |DE|^2 |CE|^2 = 4|CE|^2$, or simply $ |DE|^2 = \frac{4|CE|^2}{|CE|^2 + 4} $.

Next, we use another geometry idea that triangle area is the same no matter how it is calculated, in particular, we know $ |CD||DE| = |CE||DF| $, we eliminate $ |CD| $ by using the similarity relation again, $ |DF| = |DE|\frac{|CD|}{|CE|} = |DE|\frac{|DE|}{2} = \frac{|DE|^2}{2} = \frac{2|CE|^2}{|CE|^2 + 4} $

To finish up our story of finding lengths, $ |EF| $ can be found by similarity relation as well. $ \frac{EF}{DF} = \frac{2}{|CE|} $, therefore we get $ |EF| = \frac{2|DF|}{|CE|} = \frac{4|CE|}{|CE|^2 + 4} $

Now, let's relate these geometric results with the parametization problem. Consider this is a cross section view of the sphere with the plane containing the north pole, the origin and the point $ (u, v, 0) $. Note that the point $ (u, v, 0) $ is on the $ z = 0 $ plane which is the same as the center, therefore $ |CE| = 2\sqrt{u^2 + v^2} $.

Substitute these back to the lengths, we get :

$ |DF| = \frac{2|CE|^2}{|CE|^2 + 4} = \frac{2 \times 4(u^2 + v^2)}{4(u^2 + v^2) + 4} = \frac{2u^2 + 2v^2}{u^2 + v^2 + 1} $

$ |EF| = \frac{4|CE|}{|CE|^2 + 4} = \frac{4 \times 2\sqrt{u^2 + v^2}}{4(u^2 + v^2) + 4} = \frac{2\sqrt{u^2 + v^2}}{u^2 + v^2 + 1} $

Note the similarity of the formula with the given parametization! To finish the story, we note that by a simply coordinate transformation, $ -1 + |EF| = z $, so $ z = \frac{2u^2 + 2v^2}{u^2 + v^2 + 1} - 1 = \frac{u^2 + v^2 - 1}{u^2 + v^2 + 1} $.

$ x $ and $ y $ are slightly more complicated, we know the radial distance of the projection of the $ x $ and $ y $ point, but we do not know the angle yet. Geometrically, the angle, must be the same as the $ (u, v, 0) $ make with the origin, so we can claim that

$ x = \frac{2u}{u^2 + v^2 + 1} $ and $ y = \frac{2v}{u^2 + v^2 + 1} $ as this choice of coordinate fits the angle requirement and the radical distance requirements.

In comparison, part (b) is a lot simpler. Just substitute $ t = 0 $ and $ t = 1 $ to the given parametization will yield the two required points.

For part (c), we substitute the line into the sphere to find intersections.

$ \begin{eqnarray*} x^2 + y^2 + z^2 &=& 1 \\ (tu)^2 + (tv)^2 + (1 - t)^2 &=& 1 \\ (tu)^2 + (tv)^2 + (1 - 2t + t^2) &=& 1 \\ (tu)^2 + (tv)^2 - 2t + t^2 &=& 0 \\ (u^2 + v^2 + 1)t^2 - 2t &=& 0 \\ ((u^2 + v^2 + 1)t - 2)t &=& 0 \\ t &=& 0 \text{ or } \frac{2}{u^2 + v^2 + 1} \end{eqnarray*} $

That easily yield the parametization we wanted by substitute $ t $ back to the formula:

$ x = tu = \frac{2u}{u^2 + v^2 + 1} $

$ y = tv = \frac{2v}{u^2 + v^2 + 1} $

$ z = 1 - t = 1 - \frac{2}{u^2 + v^2 + 1} = \frac{u^2 + v^2 - 1}{u^2 + v^2 + 1} $

Story learnt? Solving things geometrically can be much harder.