Iranian Mathematical Olympiad

First, note that if $(x_1,y_1)$ and $(x_2,y_2)$ are two points with rational coordinates, then the equation of the line passing through them in the cartesian coordinates is $$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}.$$ Therefore, the equation of this line can be written using only rational numbers. We will call such lines a rational line. It is easy to see that the intersection point of two rational lines has rational coordinates. Furthermore, the ratio of the lengths of two line segments with rational end points is a rational number. Consequently, every four collinear points with rational coordinates have a rational cross ratio.

Assume to the contrary that $A_1{A}_2{A}_3{A}_4{A}_5$ is a rational pentagon that at least four lines from the lines $A_i{B}_i$ are concurrent. Without loss of generality, we can assume these four lines are $A_i{B}_i$ for $1\le i\le 4$. We will use the notation given in the following figure.



Let $\lambda =(B_1{A}_3,A_2{B}_4)$. According to the figure we have $$\begin{array}{rlrl}\lambda =(B_1{A}_3,A_2{B}_4)& =(B_1{B}_4,A_{2}U)& & (\text{Projection from }O)\\ \lambda =(B_1{A}_3,A_2{B}_4)& =(A_{5}Z,A_1{B}_4)& & (\text{Projection from }{B}_3)\\ & =(U{A}_3,B_1{B}_4)& & (\text{Projection from }O)\\ & =(B_1{B}_4,U{A}_3)& & \end{array}$$ Using the properties of cross ratio we get $${\lambda }^2=(B_1{B}_4,A_{2}U)(B_1{B}_4,U{A}_3)=(B_1{B}_4,A_2{A}_3)=\frac{\lambda }{\lambda -1}$$ Hence $\lambda$ is a root of the equation $x^2=x+1$ that has no rational root. We assumed that points $A_i$, $1\le i\le 5$, have rational coordinates. Therefore, all the lines and points in the figure are rational. Furthermore, every four collinear points with rational coordinates have a rational cross ratio. Consequently, $\lambda =(B_1{A}_3,A_2{B}_4)$ must be rational which is a contradiction since ${\lambda }^2=\lambda +1$. Therefore, at most three of the lines $A_i{B}_i$ are concurrent.