The following problem is from the 18th Balkan Mathematics Olympiad.

"In a pentagon all interior angles are congruent and all its sides have rational lengths. Prove that this pentagon is regular."

Besides the fact that no generality is lost replacing rational sides by integer sides, I am totally lost on this one. I would like hints only, as small as you can make them. I would like to come to a solution on my own if possible.

  • 2
    $\begingroup$ I'd first rewrite the premise as a statement about the complex number $\zeta : = e^{(2 \pi i)/5}$. $\endgroup$ Mar 25, 2014 at 20:17
  • $\begingroup$ BMO problems are classics. But if you really want them, I think there will surely be solutions on the internet somewhere. $\endgroup$
    – Sawarnik
    Mar 25, 2014 at 20:32

1 Answer 1


I guess I'd try representing the space spanned by unit vectors in the five relevant directions as a $\mathbb Q$-vectorspace. Its dimension should be $4$, so there exists a non-trivial linear combination which evaluates to zero (i.e. which closes the pentagon), and the coefficients for that linear combination should be all equal, corresponding to unit length.

Detais (hidden inside a spoiler block, so move mouse over to read):

Suppose your coordinate system is aligned in such a way that one of the edges is oriented with the $x$ axis. Then vectors of unit length in your five edge directions will be: \begin{align*}d_1&=\begin{pmatrix}1\\0\end{pmatrix}&d_2&=\begin{pmatrix}\cos72°\\\sin72°\end{pmatrix}=\begin{pmatrix}\frac{\sqrt5-1}4\\[1ex]\sqrt{\frac{5+\sqrt5}8}\end{pmatrix}&d_3&=\begin{pmatrix}\cos144°\\\sin144°\end{pmatrix}=\begin{pmatrix}\frac{-\sqrt5-1}4\\[1ex]\sqrt{\frac{5-\sqrt5}8}\end{pmatrix}\\&&d_5&=\begin{pmatrix}\cos72°\\-\sin72°\end{pmatrix}&d_4&=\begin{pmatrix}\cos144°\\-\sin144°\end{pmatrix}\end{align*} Now you can choose a basis for your four-dimensional $\mathbb Q$-vectorspace. One possible choice would be the following: \begin{align*}b_1 &= \begin{pmatrix}\frac14\\[1ex]0\end{pmatrix} & b_2 &= \begin{pmatrix}\frac{\sqrt5}4\\[1ex]0\end{pmatrix} & b_3 &= \begin{pmatrix}0\\[1ex]\sin72°\end{pmatrix} & b_4 &= \begin{pmatrix}0\\[1ex]\sin144°\end{pmatrix}\end{align*} To show that they are indeed a basis, you have to show that they are linearily independent over $\mathbb Q$. In other words, their coordinates must be incommensurable: \begin{equation*}\frac{\;\frac{\sqrt5}4\;}{\frac14} = \sqrt5 \not\in\mathbb Q\qquad\frac{\sin72°}{\sin144°} = \sqrt{\frac{5+\sqrt5}{5-\sqrt5}} = \sqrt{\frac{3+\sqrt5}{2}} = \frac{1+\sqrt5}{2} = \varphi \not\in\mathbb Q\end{equation*} So they are indeed independent. Now we can write our original direction vectors $d_i$ in terms of this basis, as the columns of the following matrix: \begin{equation*} M = \begin{pmatrix} 4 & -1 & -1 & -1 & -1 \\ 0 & 1 & -1 & -1 & 1 \\ 0 & 1 & 0 & 0 & -1 \\ 0 & 0 & 1 & -1 & 0 \end{pmatrix} \qquad \ker(M) = \operatorname{span}\begin{pmatrix}1\\1\\1\\1\\1\end{pmatrix} \end{equation*} So the only way to add rational multiples of your direction vectors in such a way that that the sum is zero (i.e. the polygon they describe actually closes) is by taking the same length in each of the directions. Which results in either a regular pentagon or a regular pentagram. Since the problem statement spoke about a pentagon, I assume it meant a simple polygon without self-intersection. So that rules out the pentagram case.


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