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.