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Let $p_{1},p_{2},p_{3},p_{4},p_{5}$ be five points in the euclidean plane such that no set of three of those points lie on the same line. It is easy to prove that there exists a unique pentagon such that $p_{1},p_{2},p_{3},p_{4},p_{5}$ are the midpoints of its sides (In fact there is a more general result saying that the same is true for any odd number n of points as the midpoints of the sides of an n-gon). The proof uses $\mathbb{C}$ as a model of the euclidean plane and then proves, that the system of linear equations $$\frac{1}{2}(x_{i}+x_{i+1}) = p_{i} \space\space\space \space 1 \leq i \leq 5$$ where $x_{6} = x_{1}$, has unique solutions for $x_{1},x_{2},x_{3},x_{4},x_{5}\in\mathbb{C}$ since the corresponding 5x5 matrix is invertible.

My question is whether there is a way to construct the solution using ruler and compass. (which is possible in the case with only 3 points)

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When solved, the $x_i$ are linear functions of the given $p_i$ with rational coefficients. So (theoretically) the $x_i$ should be constructible from the $p_i$. – coffeemath Feb 15 '13 at 16:36
up vote 1 down vote accepted

Consider the transformation on the plane consisting of reflecting a point over $p_1$, then $p_2$, then $p_3,p_4,p_5$. Since it is the composition of $5$ $\Pi$-rotations, it itself is a $\Pi$-rotation or just a reflection. Thus it has a fixed point $A$, which is precisely the vertex adjacent to $p_1$ and $p_5$. To find the fixed point of the transformation, apply it to any other point $X$ in the plane to get a new point $X'$. Since this was a reflection over the fixed point $A$, $A$ is simply the midpoint of $XX'$, so we can construct $A$ and all other vertices similarly.

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Thanks so much for answering this! – Vincent Pfenninger Oct 15 '13 at 17:12

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