Find a regular equilateral and equiangular 16-sided polygon that has vertices that are lattice points.
- What is the minimum side length?
- Find the vertices of such a polygon.
- Find the area of the polygon you found in (2).
I do know that this polygon exists because my teacher said that one did. I have tried to algebraically solve this problem by assigning on vertex to be (a, 0) and math the rest. I also assigned the midpoint of one side to be (a, 0). Neither have worked. It can be any type of lattice.
$\endgroup$ 152 Answers
$\begingroup$Let $L$ be a lattice and let $v_1,\ldots,v_{16}\in L$ be the vertices of such a regular $16$-sided polygon of minimum side length $m$, in order, so that $|v_{i+1}-v_i|=m$ for all $i$. Because $L$ is a lattice also $v_{i+1}-v_i\in L$, and the points $v_{i+1}-v_i$ together form a regular $16$-sided polygon. The diameter of this polygon is $2m$, so its side length is strictly smaller than $m$, contradicting the minimality of $m$. Hence no such polygon or lattice exists.
This idea is illustrated quite clearly for the regular octagon in this animation.
$\endgroup$ 4 $\begingroup$We will prove that such a polygon cannot exist by contradiction. Let $A_{1}A_{2}...A_{16}$ be such a polygon with the smallest possible side length. Then, consider the construction formed by connecting each vertex to the vertex $3$ sides away (see picture). This construction forms a smaller $16$-gon in the center, titled $B_{1}B_{2}...B_{16}$ (again, see picture). However, note that $B_{1}A_{16}A_{1}A_{2}$ is a parallelogram, so $B_{1}$ is a lattice point, and similarly for $B_{2}$...$B_{16}$. However, this means that $B_{1}B_{2}...B_{16}$ is a lattice $16$-gon with a smaller side length than $A_{1}A_{2}...A_{16}$, contradicting our earlier assumption. Thus, $\boxed{\text{no lattice } 16\text{-gon can exist.}}$
I can only conclude that your math teacher was either mistaken or trying to get you to prove that a lattice $16$-gon cannot exist. By the way, this proof applies to all regular polygons larger than a square.
