Graphing a regular pentagon

Question

I just realized that I didn't know how to graph a regular pentagon with integer coordinates... What are some possible coordinates for a regular pentagon with the uppermost point at coordinate (0,0)?

Answer 1

Suppose this were possible. Consider such a pentagon with minimal side-length $\sqrt c$. (There must be a well-defined $c$, as it is an integer, as user170231's answer implies.)

Let the pentagon be $P_0P_1P_2P_3P_4$ with $P_i=(x_i, y_i)$. First, consider the side $P_0P_1$. Its length is $$c=(x_0-x_1)^2+(y_0-y_1)^2.$$

Suppose $x_0-x_1$ and $y_0-y_1$ are both even. Then $c=0\mod 4$.

Suppose $x_0-x_1$ and $y_0-y_1$ are one odd and one even. Then $c=1\mod 4$.

Suppose $x_0-x_1$ and $y_0-y_1$ are both odd. Then $c=2\mod 4$.

Thus what $c$ is, modulo 4, constrains the parities of $x_0-x_1$ and $y_0-y_1$, and likewise all around the pentagon, because all its sides have the same length.

If $c=0\mod 4$, then all the $x_i$ have the same parity, and all the $y_i$ have the same parity, so we could translate the pentagon so that $P_0$ goes to $(0, 0)$, then halve all the coordinates, to get a smaller pentagon. This contradicts minimality of $c$.

If $c=1\mod 4$, then the $x_i-y_i$ alternate parity. And if $c=2\mod 4$, then the $x_i$ alternate parity. But there are 5 such numbers (an odd number), so this is not possible.

So there is no equilateral pentagon where both $x$ and $y$ coordinates of all its vertices are integers. (Regardless of its angles.) The same goes for any equilateral $n$-gon where $n$ is odd.

Which is similar to Calvin Lin's proof that it's not possible for both $x$ and $y$ coordinates of all its vertices to be rational. (Multiply by a common multiple of the denominators, then proceed as above.)

Answer 2

It's not possible to have integer coordinates for the vertices of a regular pentagon. Take two adjacent vertices, $(0,0)$ and $(a,b)$. Draw a line segment from $(a,b)$ to $(a,0)$, so that you have a right triangle with legs of length $a$ and $b$ and hypotenuse that coincides with the side length of the pentagon.

Since the sum of exterior angles of any $n$-gon is $360^\circ$, you have that the angle made by the hypotenuse and the $x$ axis is $\frac{360^\circ}{5}=72^\circ$, which means the remaining angle must be $18^\circ$.

By the law of sines (and skipping some work along the way), you have $$\frac{\sin18^\circ}{a}=\frac{\sin72^\circ}{b}\implies\frac{a}{b}=\frac{\sqrt5-1}{\sqrt{2(5+\sqrt5)}}$$ But this number is not a rational number, which means $(a,b)$ cannot be comprised of integer components.