# Generalizing Pythagorean Theorem: equations defining edges of a (convex) $n$-gon with $n-2$ prescribed angles?

Let $$n\geq 3$$ be an integer and $$0<\alpha_1, \dots ,\alpha_{n-2}<1$$. Let's say a tuple of positive numbers $$(e_1,\dots, e_n)$$ is nice if there is a convex $$n$$-gon $$A_1\dots A_n$$ such that $$\hat A_i=\pi\alpha_i$$ and edge lengths $$\overline{A_iA_{i+1}}=e_i$$.

(The convexity condition probably will make things rather more complicated, so I will be happy to drop that)

Pythagorean theorem says that with $$n=3$$ and $$\alpha_1=1/2$$, the nice tuples are precisely the intersection of the variety $$e_2^2=e_1^2+e_3^2$$ with the positive orthant. Of course similar result holds for any $$\alpha_1$$ and $$n=3$$.

Question: what about $$n>3$$? My guess is the set of nice tuples is still the intersection of an algebraic scheme with the positive orthant (or some polytope in it). If so, what is known about the equations defining such schemes (obviously they are homogenous, but what about things like degrees)? Any reference for such kind of problems?

(Update Apr 2021): I asked Hendrik Lenstra and he confirmed that the statement mentioned by Henri Cohen below appeared as a Proposition in his thesis (which unfortunately he has no copies left). Also according to Lenstra, it might have appeared in Nieuw Archief voor Wiskunde, but quite probably in Dutch, and I have not been able to track it down.

• This might be related to the series of papers by Ben Howard, John Millson, Andrew Snowden, and Ravi Vakil? Commented Mar 26, 2021 at 22:42
• The equation $\hat A_i=\pi\alpha_i$ is only for $1\le i\le n-2$, right? And for the edge lengths, $A_{n+1}$ is taken to mean $A_1$? Commented Mar 27, 2021 at 0:01
• @GerryMyerson: yes. Commented Mar 27, 2021 at 0:50

One has $$\overrightarrow{A_nA_1} + \cdots + \overrightarrow{A_{n-2}A_{n-1}} = \overrightarrow{A_nA_{n-1}}.$$ Taking the squared norm of both sides one arrives at $$\sum_{i=1}^{n-1} e_{i-1}^2 + 2 \sum_{1\le i < j\le n-1}e_{i-1}e_{j-1}\cos\sum_{k=i}^{j-1}(\pi-\alpha_i) = e_{n-1}^2,$$ which is the equation we are looking for.

For $$n=4$$ I studied this equation some years ago from a different viewpoint: given the side lengths of a quadrilateral, how are two of its adjacent angles related? The above equation answers this question. It turns out that it can be rewritten as follows: $$a_1^2a_2^2x_{22} + a_1^2x_{20} + a_2^2x_{02} + 2a_1a_2x_{11} + x_{00} = 0,$$ where $$\begin{gather*} a_1=\cot\frac{\alpha_1}2, \quad a_2 = \cot\frac{\alpha_2}2,\\ x_{22} = (e_1-e_2-e_3-e_4)(e_1-e_2+e_3-e_4),\\ x_{20} = (e_1+e_2+e_3-e_4)(e_1+e_2-e_3-e_4),\\ x_{02} = (e_1-e_2+e_3-e_4)(e_1-e_2-e_3+e_4),\\ x_{11} = -4e_2e_4,\\ x_{00} = (e_1+e_2-e_3+e_4)(e_1+e_2+e_3+e_4). \end{gather*}$$

• Thank you for the answer. How do we show that just that equation suffices? Commented Feb 11, 2023 at 18:27
• It suffices because the left hand side is the distance between the endpoints of a polygonal line determined by its edge lengths and angles between the edges. So if the equation holds, this open polygonal line can be closed by an edge of length $e_{n-1}$. Commented Feb 12, 2023 at 6:01

Not an answer but related: This reminds me of a nice exercise of Hendrik Lenstra: if $$P$$ is a polygon (not necessarily convex) with all edge lengths equal and all but two consecutive (important!) angles rational multiples of $$\pi$$, then so are the last two.

• That sounds very interesting, do you have a reference for it? Commented Mar 26, 2021 at 22:33