Is there a regular pentagon with a rational point on each edge?

Question

This question was asked by Yaakov Baruch in the comments to the question Can a regular icosahedron contain a rational point on each face? It seems that this question deserves special attention.

Answer 1

I'm considering a more generic problem, allowing points to lie on lines containing edges. Let use the $\mathbf p_i$ for the pentagon vertices and $\mathbf q_i$ for the rational points. The pentagon is fixed by its center $\mathbf c$ and $\mathbf r = \mathbf p_1 - \mathbf c$. Let $R$ be the rotation matrix by $\frac{2\pi}{5}$. Also let $T = R^{-1} = R^\top$.

The vertices of the pentagon are given by $$ \mathbf p_i = \mathbf c + R^i \mathbf r, \quad i = 0,\dots,4. $$ Index $i$ in $\mathbf p_i$ is considered modulo 5.

The points $\mathbf q_i$ can be represented as $$ \mathbf q_i = (1 - \lambda_i)\mathbf p_i + \lambda_i \mathbf p_{i+1} = \mathbf c + R^i((1 - \lambda_i) I + \lambda_i R) \mathbf r, \quad \lambda_i \in \mathbb R. $$ We do not restrict $\lambda_i \in [0, 1]$, so $\mathbf q_i$ can be anywhere on the line passing through $\mathbf p_i$ and $\mathbf p_{i+1}$.

Let's consider auxiliary points $\mathbf a_i$ defined as $$ \mathbf a_i = T^i (\mathbf q_i - \mathbf c). $$ This points satisfy $\mathbf a_i = (1 - \lambda_i) \mathbf r + \lambda_i R \mathbf r$ so they all lie in some line $\ell = \operatorname{aff}(\mathbf r, R \mathbf r)$.

Let's construct such linear combinations of $\mathbf a_i$ so they are free of $\mathbf c$. This basically reduces to finding trivial combinations $\sum_{k=0}^4 \alpha_k T^k = O$. It's easy to see that columns of the linear combination vanish at the same time, so the equality reduces to $$ \sum_{k=0}^4 \alpha_k \cos \frac{2\pi k}{5} = 0\\ \sum_{k=0}^4 \alpha_k \sin \frac{2\pi k}{5} = 0 $$ There are three linearly independent solutions to the system, forming the following null space matrix $$ \begin{pmatrix} \phi \\ -1 & \phi\\ \phi & -1 & \phi\\ & \phi & -1\\ && \phi \end{pmatrix}. $$ Here $\phi = \frac{1 + \sqrt{5}}{2}$, the golden ratio. It is easy to verify that $$\phi T^{i+1} - T^i + \phi T^{i-1} = T^i (\phi (R + T) - I) = O.$$

The combinations that are free from $\mathbf c$ are: $$ \mathbf b_i = \phi \mathbf a_{i-1} - \mathbf a_i + \phi \mathbf a_{i+1} = \phi T^{i-1}\mathbf q_{i-1} - T^i\mathbf q_i + \phi T^{i+1}\mathbf q_{i+1}, \quad i = 1, 2, 3. $$ Points $\mathbf b_i$ also lie on a same line $\ell' = (2\phi - 1) \ell$. Therefore $$ (\mathbf b_3 - \mathbf b_2) \vee (\mathbf b_2 - \mathbf b_1) = 0. $$ Here $\vee$ is the pseudoscalar (or skew) product.

By using the trigonometric form $T^k = \cos \frac{2\pi k}{5} I + \sin \frac{2\pi k}{5} J$ and performing some lengthy computation, one can obtain $$ \frac {4}{\phi^2} (\mathbf b_3 - \mathbf b_2) \vee (\mathbf b_2 - \mathbf b_1) = \\ = \left(\sum_{i=0}^4 \mathbf q_i \vee \mathbf q_{i-1}\right) (\sqrt{5} - 1) + \left(\sum_{i=0}^4 \mathbf q_i \vee \mathbf q_{i-2}\right) (\sqrt{5} + 3) + \\ + \left(\sum_{i=0}^4 \mathbf q_i \cdot (\mathbf q_{i-1}-\mathbf q_{i-2})\right) \sqrt{10 + 2\sqrt{5}}. $$ The indices are considered modulo 5.

Now let $$ A = \sum_{i=0}^4 \mathbf q_i \vee \mathbf q_{i-1},\\ B = \sum_{i=0}^4 \mathbf q_i \vee \mathbf q_{i-2},\\ C = \sum_{i=0}^4 \mathbf q_i \cdot (\mathbf q_{i-1}-\mathbf q_{i-2}),\\ D = A + B,\\ E = 3B - A. $$ Note that $A, B, C, D, E \in \mathbb Q$.

We finally have a necessary condition $$ D \sqrt{5} + E + C \sqrt{10 + 2\sqrt{5}} = 0. $$ It is easy to show that there is no nontrivial solution: $$ 5D^2 + E^2 + 2\sqrt{5}DE = C^2 (10 + 2\sqrt{5})\\ 5D^2 + E^2 - 10C^2 = 2(C^2 - DE)\sqrt{5}\\ C^2 - DE = 0\\ 5D^2 + E^2 = 10C^2 = 10DE\\ (E - 5D)^2 = 20D^2 \\ E = (5 \pm 2\sqrt{5})D\\ C = D = E = 0. $$ So the only possibility is $A = B = 0$. But $A$ is twice the oriented area of the polygon formed by $\mathbf q_{0},\dots,\mathbf q_{4}$. Similarly, $B$ is twice the area of the star-shaped polygon with permuted vertices $\mathbf q_0, \mathbf q_2, \mathbf q_4, \mathbf q_1, \mathbf q_3$.

This surely can't happen when $\mathbf q_{i}$ lie on the sides, since then polygon $\mathbf q_0, \mathbf q_1, \mathbf q_2, \mathbf q_3, \mathbf q_4$ is convex and nonempty.

Update. However there are rational solutions to $A = B = C = 0$, for example $$ \mathbf q_0 = \left(\frac{1}{5},\frac{1}{2}\right), \mathbf q_1 = (1,2), \mathbf q_2 = \left(1,\frac{1}{2}\right), \mathbf q_3 = \left(2,\frac{1}{2}\right), \mathbf q_4 = \left(2,\frac{7}{6}\right). $$ This particular one was obtained by trial and error method by plugging 1 and 2 for unknowns and solving the remaining system.

Knowing $\mathbf q_i$ we can compute $\mathbf b_i$, obtaining the $\ell'$ line. Divinding it by $2\phi-1$ we get the $\ell$ line where $\mathbf a_i, \mathbf r$ and $R\mathbf r$ are located. The $\mathbf r$ can be found from the following linear system $$ ((2\phi-1) \mathbf r - \mathbf b_1) \vee (\mathbf b_2 - \mathbf b_1) = 0\\ ((2\phi-1) R\mathbf r - \mathbf b_1) \vee (\mathbf b_2 - \mathbf b_1) = 0 $$ Similarly $\mathbf c$ can be obtained as the solution to $$ ((2\phi-1) (\mathbf q_0 - \mathbf c) - \mathbf b_1) \vee (\mathbf b_2 - \mathbf b_1) = 0\\ ((2\phi-1) T (\mathbf q_0 - \mathbf c) - \mathbf b_1) \vee (\mathbf b_2 - \mathbf b_1) = 0 $$

The layout of the points and the pentagon is shown in the picture below:

Answer 2

Update 3: The answer by uranix which finally settles the question is extremely clever! The essential point is the algebraic relation between $C, D, E$. The following Sage code (which can be run e.g. in the SageMathCell) verifies that his calculation is indeed correct:

vars = [f'l_{i}' for i in range(5)]
vars += ['cx', 'cy', 'rx', 'ry', 's']
S = PolynomialRing(QQ, vars)
S.inject_variables()
f = s^4 - 5/4*s^2 + 5/16 # s = sin(2*pi/5), f = minpol of s
c = 2*s^2 - 3/2          # c = cos(2*pi/5)

R = matrix([[c, s], [-s, c]])

def redvec(v):
    return vector((v[0]%f, v[1]%f))

def p(i):
    return redvec(vector(S, (cx, cy)) + R^(i%5)*vector(S, (rx, ry)))

def l(i):
    return S.gen(i%5)

def q(i):
    return (1-l(i))*p(i) + l(i)*p(i+1)

def pseu(u, v):
    return matrix([u,v]).det()

def scalp(u, v):
    return u[0]*v[0] + u[1]*v[1]

A = sum(pseu(q(i), q(i-1)) for i in range(5))
B = sum(pseu(q(i), q(i-2)) for i in range(5))
C = sum(scalp(q(i), q(i-1)-q(i-2)) for i in range(5))
D = A + B
E = 3*B - A

uranix = ((5*D^2 + E^2 -10*C^2)^2 - 20*(C^2 - D*E)^2) % f
print(uranix == 0)

Update 2: (Answering the question of uranix in the comments) In L, we set $m=m_0+m_1s+m_2s^2+m_3s^3$, and grouping by powers of $s$, we get $16$ equations in $y1, x2, y2, x3, y3, x4, y4$ and the $m_i$'s. After running the displayed code from the original answer, run the following code:

vars += ['m0', 'm1', 'm2', 'm3']
S = PolynomialRing(QQ, vars)
S.inject_variables()
LL = [S(z).subs({m:m0 + m1*s + m2*s^2 + m3*s^3})%f for z in L]
N = []
for z in LL:
    N += [S(_) for _ in z.polynomial(s).coefficients()]
I = ideal(N)
cubic = 64*m0*m1^2 - 64*m0^2*m2 - 80*m0*m2^2 - 20*m2^3 + 160*m0*m1*m3 + 40*m1*m2*m3 + 80*m0*m3^2 + 25*m2*m3^2 - 64*m2
print(cubic in I)
print(I.elimination_ideal(S.gens()[:7]))

The second to last line verifies that the cubic indeed has to vanish. And the last line shows how the cubic was obtained from eliminating variables. (If one would like to know how cubic is expressed in term of the polynomials in N, run cubic.lift(I).)

Update 1: With the notation from below, an example where an edge has slope $m$ requires $m\in\mathbb Q(s)$ where $s=\sin(2\pi/5)$. Note that $s$ has degree $4$ over $\mathbb Q$. If we write $m=m_0+m_1s+m_2s^2+m_3s^3$ with rational $m_i$, then a further necessary condition is \begin{equation} 64m_0m_1^2 - 64m_0^2m_2 - 80m_0m_2^2 - 20m_2^3 + 160m_0m_1m_3 + 40m_1m_2m_3 + 80m_0m_3^2 + 25m_2m_3^2 - 64m_2=0. \end{equation} This condition is essentially sufficient for the following: There are $5$ distinct rational points $u_i$, one on each line through an edge. This cubic has many rational solutions. (I guess it is rationally parametrized or at least unirational) Still, none of these points so far forced all the $u_i$'s to lie inside each edge. This additional condition amounts to high degree polynomial inequalities.

---

This is an attempt giving some partial results.

Set $s=\sin(2\pi/5)$ and $c=2s^2 - 3/2=\cos(2\pi/5)$. We show that if there is a positive answer, then the slopes of the edges are contained in the field $\mathbb Q(s)$, but none of the slopes is rational.

Let $u_0,\ldots,u_4$ be five putative rational points on the five sides of the regular pentagon. We allow that some points coincide. Note however that $u_i=u_j$ can only happen if $i-j\in\{-1,0,1\}$ (indices taken modulo $5$).

The rational group $\text{SO}(2,\mathbb Q)$ of rotations is dense in $\text{SO}(2,\mathbb R)$ (in view of the rational parametrization $x=2t/(1+t^2), y=(1-t^2)/(1+t^2)$ of the unit circle).

In particular, we may and do assume the following:

• None of the slopes is vertical.
• If $(x_i,y_i)=u_i\ne u_j=(x_j,y_j)$, then $x_i\ne x_j$ and $y_i\ne y_j$.
• $x_0=y_0=0$, $x_1=1$.

Let $m$ be the slope of the edge through $u_0$. By repeated rotation of the vector $(1,m)$ by $2\pi/5$, we get the slopes through the $u_i$'s. By intersecting the lines through these points, we get the vertices of the pentagon. Let $p_i$ be the intersection of the lines through $u_i$ and $u_{i+1}$. The condition which has to hold is that the vector $p_{i+2}-p_{i+1}$ is $p_{i+1}-p_i$ rotated by $2\pi/5$.

For $i=3$ this yields the condition \begin{align} q_0+q_1m &= 0\text{ where}\\ q_0 &= (-4x_3 + 4x_4)s^3 + (2y_1 + 2y_3)s^2 + (3x_3 - 2x_4 - 1)s - 3/2y_1 - y_3 + 1/2y_4\\ q_1 &= (-4y_3 + 4y_4)s^3 + (-2x_3 - 2)s^2 + (-y_1 + 3y_3 - 2y_4)s + x_3 - 1/2x_4 + 3/2 \end{align} We use that $s$ has degree $4$ over $\mathbb Q$, and $s^4 - 5/4s^2 + 5/16=0$.

Suppose that $m$ is not in the field generated by $s$. Then $q_1=q_0=0$. The linear independence of $1,s,s^2,s^3$ yields $8$ equations in the $x_i$'s and $y_i$'s. Let $q_i[j]$ be the coefficient in $q_i$ of $s^j$. We get the contradiction \begin{equation} 0 = q_0[3] + 8q_1[0] + 2q_1[2] = (-4x_3 + 4x_4)+8(x_3 - 1/2x_4 + 3/2)+2(-2x_3 - 2)=8. \end{equation} Similarly, we show that $m$ is not rational. Suppose that $m$ is rational. Then $q[j]:=q_0[j]+mq_1[j]$ vanishes for $j=0,1,2,3$. One computes that \begin{equation} 8mq[0]-4q[1]+4mq[2]-3q[3]=4(1-x_4)(1+m^2), \end{equation} so $4(1-x_4)(1+m^2)=0$. Of course $1+m^2\ne0$, hence $x_4=1=x_1$, which contradicts our assumption $u_4\ne u_1$ (and different $x$-coordinates for different $u_i$'s).

The SageMath code below does the actual computation. In order to continue, one can use $q$ to eliminate $m$ in the other equations (collected in the list L in the program). Using Groebner bases one gets certain conditions on the $x_i$'s and $y_i$'s, but I wasn't able to get a contradiction, nor a positive example. The computations just became too heavy, so a better idea is needed.

vars = [f'x_{i}' for i in range(2, 5)]
vars += [f'y_{i}' for i in range(1, 5)]
vars += ['m', 's']
R = PolynomialRing(QQ, vars)
R.inject_variables()
f = s^4 - 5/4*s^2 + 5/16
c = 2*s^2 - 3/2

def u(i):
    i = i%5
    if i == 0:
        return 0, 0
    if i == 1:
        return 1, y_1
    return R.gen(i-2), R.gen(i+2)

def rot(u):
    """Rotate vector u by 2*pi/5"""
    x, y = u
    return (x*c - y*s)%f, (x*s + y*c)%f

def diff(u, v):
    return u[0]-v[0], u[1]-v[1]

slope = (1, m)
p = []
for k in range(7):
    ux, uy = u(k)
    sux, suy = slope
    slope = rot(slope)
    vx, vy = u(k+1)
    svx, svy = slope
    x = suy*svx*ux - sux*svx*uy - sux*svy*vx + sux*svx*vy
    y = suy*svy*ux - sux*svy*uy - suy*svy*vx + suy*svx*vy
    p.append((x%f, y%f))

deltas = [diff(p[k+1], p[k]) for k in range(6)]

L = [diff(deltas[k+1], rot(deltas[k]))[0] for k in range(4)]
q = [z for z in L if z.degree(m) == 1][0]
lq = [R(z) for z in q.polynomial(m)]

def L2N(L): #Get the coefficients of the polynomial in s
    N = []
    for z in L:
        N += [R(_) for _ in z.polynomial(s).coefficients()]
    return N

print('Checking that m in Q(s):')
N = L2N(lq)
print(1 in ideal(N), R(1).lift(N))

print('Checking that m is not rational:')
N = L2N([q])
w = (1-x_4)*(m^2+1)
print(w in ideal(N), w.lift(N))

Answer 3

It is possible to modify Uranix's lovely answer to show that the only $n \ge 3$ for which there is a regular $n$-gon with a rational point on each side are $n=3,4,8$ (for which simple constructions have already been demonstrated here).

Observing that one can obtain a regular $n$-gon by extending $n$ sides of a regular $nm$-gon, it suffices to prove impossibility in the case in which $n$ is $16,6,9$, or an odd prime $p > 3$. Fix such an $n$ and suppose for sake of contradiction such an $n$-gon exists. Let $q_i$ be a rational point on the $i$th side (oriented counterclockwise), $p$ the center of the polygon, and $T$ rotation by $-\frac{2 \pi}{n}$. It will also be convenient to define $c(k) = \cos\left(\frac{2\pi k}{n}\right)$ and $s(k) = \sin\left(\frac{2\pi k}{n}\right)$.

Now exactly as in Uranix's answer the points $$ T^i(q_i-p) $$ lie in a common line. Thus for each $k$, the sum $$ a_k = \sum_{i=1}^n (T^i(q_i-p)) \vee (T^{i+k}(q_{i+k}-p)) $$ is equal to $0$ (where indices are taken modulo $n$). Now using $x \vee T^ky = c(k) x \vee y - s(k) x \cdot y$ we have \begin{align*} a_k &= \sum_{i=1}^n (q_i-p) \vee (T^{k}(q_{i+k}-p))\\ &= \sum_{i=1}^n [c(k) (q_{i}-p) \vee (q_{i+k}-p) - s(k) (q_{i}-p) \cdot (q_{i+k}-p)]\\ &= c(k)\sum_{i=1}^n q_{i} \vee q_{i+k} - s(k) \sum_{i=1}^n q_{i}\cdot q_{i+k}+2s(k)\sum_{i=1}^n p\cdot (q_i - p ).\\ \end{align*} We wish to be rid of those terms depending on $p$. Denote $V_k = \sum_{i=1}^n q_{i} \vee q_{i+k}$ and $D_k = \sum_{i=1}^n q_{i} \cdot q_{i+k}$, and observe that \begin{align*} 0 &= 2s(m)a_k - 2s(k)a_m \\&= 2s(m)c(k)V_k - 2s(k)c(m)V_m-2s(m)s(k)(D_m-D_k) \\&= s(m+k)(V_k-V_m)+s(m-k)(V_k+V_m)-(c(m-k)-c(m+k))(D_m-D_k) \end{align*} This is finally our necessary condition for solutions to exist. Most of the argument works with $m=2,k=1$ where we have \begin{equation} \label{eqn:main} s(3)(V_1-V_2)+s(1)(V_1+V_2)-(c(1)-c(3))(D_2-D_1)=0\tag{$*$} \end{equation} Note the crucial facts that $D_k,V_k$ are rational and that $V_1$ and $V_2$ are the volumes of the polytopes $q_1q_2q_3...$ and $q_1q_3q_5...$ and thus positive when $n > 4$.

We now derive a contradiction for every listed $n$.

Firstly if $n$ is an odd prime and greater than $3$, we have that $\{s(1),s(3),c(1),c(3)\}$ are $\mathbb Q$-linearly independent since $\{e^{\pm\frac{2\pi i}{n}},e^{\pm 3\frac{2\pi i}{n}}\}$ are $\mathbb Q(i)$-linearly independent. Thus for \eqref{eqn:main} to hold we require $$(V_1-V_2)=(V_1+V_2)=(D_2-D_1)=0$$ so in particular $V_1=V_2=0$, contradicting positivity of the $V_k$.

If $n=6$, then $s(3)$ vanishes and \eqref{eqn:main} becomes $$ 0=\frac{\sqrt{3}}{2}(V_1+V_2)-\frac{3}{2}(D_2-D_1), $$ so again $V_1+V_2=0$, contradicting the positivity of the $V_k$.

If $n=9$ we use the alternate condition $$s(5)(V_2-V_3)+s(1)(V_2+V_3)-(c(1)-c(5))(D_3-D_2)=0$$ for which the same argument as in the odd prime case applies to show $s(1), s(5), c(1)-c(5)$ are linearly independent, giving $(V_2-V_3)=(V_2+V_3)=(D_3-D_2)=0$ and thus $V_2=0$, again a contradiction.

Finally if $n=16$, then $c(1)$ and $c(3)$ are $\mathbb Q$-linearly independent, but $s(1) = c(3)$ and $s(3)=c(1)$, so \eqref{eqn:main} becomes $$ V_1-V_2 = D_2-D_1 = -V_1-V_2 $$ and thus $2V_1=0$ in this case, contradicting the positivity of the $V_k$.

This answer of course relies heavily on the stipulation that the rational points lie on each side. If this is removed there are solutions for at least the pentagon (as shown by Uranix) and the hexagon as well. I am curious if any other $n$ admit such solutions.