Can a portion of a hypocycloid be a regular polygon? Hypocycloids are curves that generally don't include straight lines. A significant exception is a hypocycloid with 2 cusps, generated by rolling one circle inside another having twice the radius of the first, which will generate a completely straight line and no curved portion. An example and a proof of this special case is shown in this link. 
Aside from this special case, I want to know if a hypocycloid can be generated such that it contains straight line segments that form the edges of a regular polygon. Specifically, I want to know if a hypocycloid of 5 cusps can be generated so that it contains a true regular pentagon, and if so, what is the "pen position" as a function of the radius of the smaller circle.
The following images were generated at Spiromath. 

This first one shows generally what I am trying to accomplish. The radius of the larger circle (A) is 120. The radius of the smaller circle (B) is 72. On the web page, it is defined as -72 to cause the smaller circle to roll inside the larger instead of outside, but the end result is still a radius of 72. The "pen position", at the end of the green line is 96 in this example, but I think a value of 106 is closer to generating the desired result. The ratio of 72/120 reduces to 3/5 and the denominator defines the number of cusps while the numerator is related to which "vertices" are "connected" and the number of revolutions required to generate the closed curve.

The second picture is the same as the first, but with several additional variations of the "pen position" which was varied by increments of 4 to make the drawing. Notice that increasing the pen position causes the portion of the hypocycloid within circle A to become more and more convex, while decreasing the pen position causes the portion within circle A to become increasingly concave. The portion of the hypocycloid outside circle A (the "points" of the star") are always convex.
Since the portions within the larger circle vary from convex to concave as the pen position is adjusted within the range shown, I am postulating that there is at least one pen position at which a significant part of curve becomes a straight line. Simply from trial and error, I am speculating that the straight part may span across the entire arc of the outer circle, or it may only span the length of the edge of the inscribed pentagon, or maybe it's somewhere in between.
That's about as far as I can take it. I'd like to see if someone can provide a formula for the "pen position" that will generate a pentagon precisely bounded by exactly straight lines, and also show how far those line actually remain straight, or if this is impossible, provide a proof that it is impossible. I am specifically looking for the case of a pentagon, but it would be great to have a general answer that applies to any hypocycloid bounding any regular polygon such as an equilateral triangle, square or other n-gon.
UPDATE: I discovered that although I have been using the term hypocycloid, the curve I am actually describing a hypotrochoid. A hypocycloid is a special instance of a hypotrochoid where the pen position equals the radius of the smaller circle. This will not be true for the solution to this question so the proper term for the curve I'm trying to find is a hypotrochoid. The parametric equations for a hypotrochoid are found at MathWorld. 
I am also adding a new image created with Geometer's Sketchpad. It shows a pair of complimentary hypotrochoids. In this case, the red radius is 3/5 of the green radius and the blue radius is 2/5 of the green radius. Their "pen positions" have been adjusted by eye until the curve closely approaches the vertical lines indicated to the left of center. The red hypotrochoid is similar to those in the other images, although the blue one is a viable alternative solution to my problem.

Finally, note that I have a good background in algebra and trig, but no analytic geometry or calculus, so answers that don't require calculus are preferred, although an answer that requires calculus is better than no answer at all, although I may need additional explanation in that case. 
Thanks for playing.
 A: The parametric equation of the spirograph curve is
$$\begin{align}
x(\theta)&=(a+b)\cos\theta+p\cos(\tfrac{a+b}{b}\,\theta), \\
y(\theta)&=(a+b)\sin\theta+p\sin(\tfrac{a+b}{b}\,\theta),
\end{align}$$
where $\theta$ is how far the center of circle B has turned counterclockwise, and $p$ is the "pen position" parameter. For your chosen values, for example, we have
$$\begin{align}
x(\theta)&=48\cos\theta - 96\cos(-\tfrac23\theta), \\
y(\theta)&=48\sin\theta - 96\sin(-\tfrac23\theta), \\
\end{align}$$
as you can see plotted on WolframAlpha.

For it to contain a vertical line, the $x$-coordinate would have to be constant for a finite interval of values of $\theta$. But $x(\theta)$ is a linear combination of trigonometric functions of $\theta$, and cannot be constant on a finite interval unless it is constant for all values of $\theta$ (which is what happens with the hypocycloid with 2 cusps). The best you can do is to have $x''(\theta)=0$ at $\theta=0$, which happens when $p=-\frac94(a+b)$, which is $-108$ in your case. Under these conditions the curve has third-order contact with a vertical line, while in all other cases it has only first-order contact.

