Drawing of a regular heptadecagon

Question

I found a very interesting blog (Gauss: Constructibility of Heptadecagon) that provides a complete demonstration of the correctness of the geometric construction of the regular heptadecagon found by Herbert William Richmond in 1893.

The demonstration is quite clear to me, except in its key point, described in step 8 of the Theorem 5 (Constructibility of Heptadecagon using compass and ruler):

«(8) Let S, Q be the points where the circle $C_3$ intersects with $CP_0$»

To understand the meaning of this sentence it is not necessary to analyze the whole demonstration, but only the seven preceding steps of theorem 5. The question is "simple": how can $C_3$ intersect $CP_0$ at the median point Q of $PP_0$ segment? (see step 5 for the definition of point Q).

Only by this result you can understand the next step 26 and complete the demonstration but something escapes me and I can not figure out how $C_3$ can intersect the $C_1$ diameter exactly in Q, as defined in step 5.

Can anyone help me to understand? Any algebraic and / or trigonometric suggestions?

Thanks in advance

Fabrizio Faraone

Answer 1

The whole point is that the Galois group of the minimal polynomial of $\cos\frac{2\pi}{17}$ over $\mathbb{Q}$ is isomorphic to $\left(\mathbb{Z}/2\mathbb{Z}\right)^4$, hence $\cos\frac{2\pi}{17}$ can be written in terms of nested square roots, leading to the following nasty expression:

$$\frac{−1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}−2\sqrt{34+2\sqrt{17}}}}{16}$$ This clearly is a constructible number, and the Euclidean approach for constructing it by intersecting the least amount of circles is not that important; one just has to recall that if two segments with lengths $a,b$ are given, $\sqrt{ab}$ can be constructed in the following way:

Now you have all the ingredients for devising your personal construction of the regular heptadecagon, and understanding the Gaussian construction as well.