UTM Ideals Varieties and Algorithm - Chapter 1 Section 2 Exercise 14

Problem:

Solution:

a)

This is obvious, the arm cannot possibly reach anything outside the circle with radius 6.

b)

This is again obvious, the node between second and third arm has to stay on the circle with radius 5, the only remaining freedom is the third arm which can bring us anywhere in the annulus.

c)

We already know we are able to reach any point in the annulus, now we only need to worry about points in the inner circle with radius 4. Suppose we fold the second arm completely backwards to be parallel with the first arm, we can reach all points with a circle with radius 2.

On the other hand, if we can fix the angle between the first arm and the second arm to be at an angle such that the node of the second arm is of distance 3 from the origin, then we are done because now the last arm can cover all points between 2 and 4 from the origin.

To do that, we note that the first arm is distance 3 from the origin, the second arm is 2 distance from the first arm tip, and the second arm tip is again distance 3, so we form an isosceles triangle and therefore the angle is simply $ \cos^{-1} \frac{1}{3} $.