Problem 1297

We will find such a pentagon. The obstruction is, symmetric pentagons (for which it could be easier to show) have always at least one pair of lines $A_iA_{i+3}$, $A_{i+1}A_{i+2}$ parallel. First we will solve more elementary problem - we will find pentagon $A_1A_2A_3A_4A_5$ with only four points $B_i$ collinear. Here we afford to look for it among symmetric pentagons. To simplify the situation, say $A_2, A_3, A_4$ be vertices of the square $QA_2A_3A_4$ with $QA_2 = 1$ and $A_1, A_5$ be on sides $QA_2$ and $QA_4$ with $QA_1 = QA_5 = p$ (fig. 3). By symmetry $B_1B_2 \parallel B_3B_5$. It can be observed when $p \to 0$, i.e. when $A_1, A_5$ be near $Q$, the

line $B_3B_5$ is more closely to $Q$ than $B_1B_2$. On the other side, when $p \to 1$, points $A_1, A_5$ be near $A_2$, $A_4$ and $B_1B_2$ is more closely to $Q$ (or even on the other side of $Q$) than $B_3B_5$. Thus we can expect for some $p \in (0, 1)$ both lines be identical and $B_1, B_2, B_3, B_5$ collinear. We will find such $p$.

Let $B_5Q = B_3Q = q$ and $B_1A_2 = r$. By similarity of triangles $B_5QA_5, B_5A_2A_3$ we get
$$\frac{q}{p} = \frac{q+1}{1} \quad \text{or} \quad q = \frac{p}{1-p}.$$
By similarity of triangles $B_1A_2A_1, B_1A_3A_4$ we get
$$\frac{r}{1-p} = \frac{r+1}{1} \quad \text{or} \quad r = \frac{1-p}{p}.$$
Finally, to reach $B_1$ lying on the line $B_3B_5$, it suffices triangles $B_5QB_3, B_5A_2B_1$ be similar. This holds when
$$\frac{q}{q} = \frac{q+1}{r} \quad \text{or} \quad q+1 = r.$$
Substituting by previous we obtain the equation
$$\frac{p}{1-p} + 1 = \frac{1-p}{p}$$
which can be transformed into $p^2 - 3p + 1 = 0$. The only solution in $(0, 1)$ is $p = (3 - \sqrt{5})/2$. For this value points $B_1, B_3, B_5, B_2$ lies on one line. Moreover lines $A_1A_5, A_2A_4$ are parallel to it. In some sense these three lines intersect "in infinity" in "point" $B_4$ and all points $B_i$ be collinear. To satisfy given conditions it suffices to find proper map, which maps "point from infinity" to particular point (and preserves all the other necessary properties, e.g. maps lines to lines etc.). Such a map is projection (fig. 4). Consider ordinary cartesian coordinate system in space. Pentagon $A_1A_2A_3A_4A_5 = U$ can be imbedded

into the plane $Oyz$ with $A_2$ be in origin and with $A_1$, $A_3$ be on positive axis $z$, $y$. Let $P \equiv (2,0,-1)$ be the projection-point. Every line $PA_i$ intersects the plane $Oxy$ in point $A'_i$. Thus we get pentagon $A'_1A'_2A'_3A'_4A'_5 = U'$. It follows from the properties of the map, that $U'$ satisfies conditions given in the formulation of the problem. To check this, it is also sufficient to do some calculation. Namely one can easily enumerate the coordinates of $A'_i$ in the plane $Oxy$ and get
$$A'_1 \equiv (3 - \sqrt{5}, 0), \quad A'_2 \equiv (0, 0), \quad A'_3 \equiv (1, 0), \quad A'_4 \equiv (1, \frac{1}{2}), \quad A'_5 \equiv (1, \frac{3-\sqrt{5}}{4}).$$
By another manual calculations it is easy to check $B'_1, B'_2, B'_3, B'_4, B'_5$ be collinear (fig. 5).